6.2: Problems on Random Variables and Probabilities
Exercise \(\PageIndex{1}\)
The following simple random variable is in canonical form:
\(X = -3.75 I_A - 1.13 I_B + 0 I_C + 2.6 I_D\).
Express the events \(\{X \in (-4, 2]\}\), \(\{X \in (0, 3]\}\), \(\{X \in (-\infty, 1]\}\), and {\(X \ge 0\)} in terms of \(A\), \(B\), \(C\), and \(D\).
- Answer
-
- \(A \bigvee B \bigvee C\)
- \(D\)
- \(A \bigvee B \bigvee C\)
- \(C\)
- \(C \bigvee D\)
Exercise \(\PageIndex{2}\)
Random variable \(X\), in canonical form, is given by \(X = -2I_{A} - I_B + I_C + 2I_D + 5I_E\).
Express the events \(\{X \in [2, 3)\}\), \(\{X \le 0\}\), \(\{X < 0\}\), \(\{|X - 2| \le 3\}\), and \(\{X^2 \ge 4\}\), in terms of \(A, B, C, D, and E\).
- Answer
-
- \(D\)
- \(A \bigvee B\)
- \(A \bigvee B\)
- \(B \bigvee C \bigvee D \bigvee E\)
- \(A \bigvee D \bigvee E\)
Exercise \(\PageIndex{3}\)
The class \(\{C_j: 1 \le j \le 10\}\) is a partition. Random variable \(X\) has values {1, 3, 2, 3, 4, 2, 1, 3, 5, 2} on \(C_1\) through \(C_{10}\), respectively. Express X\) in canonical form.
- Answer
-
T = [1 3 2 3 4 2 1 3 5 2]; [X,I] = sort(T) X = 1 1 2 2 2 3 3 3 4 5 I = 1 7 3 6 10 2 4 8 5 9
\(X = I_A + 2I_B + 3I_C + 4I_D + 5I_E\)
\(A = C_1 \bigvee C_7\), \(B = C_3 \bigvee C_6 \bigvee C_{10}\), \(C = C_2 \bigvee C_4 \bigvee C_8\), \(D = C_5\), \(E = C_9\)
Exercise \(\PageIndex{4}\)
The class \(\{C_j: 1 \le j \le 10\}\) in Exercise has respective probabilities 0.08, 0.13, 0.06, 0.09, 0.14, 0.11, 0.12, 0.07, 0.11, 0.09. Determine the distribution for \(X\)
- Answer
-
T = [1 3 2 3 4 2 1 3 5 2]; pc = 0.01*[8 13 6 9 14 11 12 7 11 9]; [X,PX] = csort(T,pc); disp([X;PX]') 1.0000 0.2000 2.0000 0.2600 3.0000 0.2900 4.0000 0.1400 5.0000 0.1100
Exercise \(\PageIndex{5}\)
A wheel is spun yielding on an equally likely basis the integers 1 through 10. Let C i be the event the wheel stops at \(i\), \(1 \le i \le 10\). Each \(P(C_i) = 0.1\). If the numbers 1, 4, or 7 turn up, the player loses ten dollars; if the numbers 2, 5, or 8 turn up, the player gains nothing; if the numbers 3, 6, or 9 turn up, the player gains ten dollars; if the number 10 turns up, the player loses one dollar. The random variable expressing the results may be expressed in primitive form as
\(X = -10I_{C_1} + 0I_{C_2} + 10I_{C_3} - 10I_{C_4} + 0I_{C_5} + 10I_{C_6} - 10I_{C_7} + 0I_{C_8} + 10I_{C_9} - I_{C_{10}}\)
- Determine the distribution for \(X\), (a) by hand, (b) using MATLAB.
- Determine \(P(X < 0)\), \(P(X > 0)\).
- Answer
-
p = 0.1*ones(1,10); c = [-10 0 10 -10 0 10 -10 0 10 -1]; [X,PX] = csort(c,p); disp([X;PX]') -10.0000 0.3000 -1.0000 0.1000 0 0.3000 10.0000 0.3000 Pneg = (X<0)*PX' Pneg = 0.4000 Ppos = (X>0)*PX' Ppos = 0.300
Exercise \(\PageIndex{6}\)
A store has eight items for sale. The prices are $3.50, $5.00, $3.50, $7.50, $5.00, $5.00, $3.50, and $7.50, respectively. A customer comes in. She purchases one of the items with probabilities 0.10, 0.15, 0.15, 0.20, 0.10 0.05, 0.10 0.15. The random variable expressing the amount of her purchase may be written
\(X = 3.5 I_{C_1} + 5.0 I_{C_2} + 3.5 I_{C_3} + 7.5 I_{C_4} + 5.0 I_{C_5} + 5.0I_{C_6} + 3.5 I_{C_7} + 7.5I_{C_8}\)
Determine the distribution for \(X\) (a) by hand, (b) using MATLAB.
- Answer
-
p = 0.01*[10 15 15 20 10 5 10 15]; c = [3.5 5 3.5 7.5 5 5 3.5 7.5]; [X,PX] = csort(c,p); disp([X;PX]') 3.5000 0.3500 5.0000 0.3000 7.5000 0.3500
Exercise \(\PageIndex{7}\)
Suppose \(X\), \(Y\) in canonical form are
\(X = 2 I_{A_1} + 3 I_{A_2} + 5 I_{A_3}\) \(Y = I_{B_1} + 2 I_{B_2} + 3I_{B_3}\)
The \(P(A_i)\) are 0.3, 0.6, 0.1, respectively, and the \(P(B_j)\) are 0.2 0.6 0.2. Each pair {\(A_i, B_j\)} is independent. Consider the random variable \(Z = X + Y\). Then \(Z = 2 + 1\) on \(A_1 B_1\), \(Z = 3 + 3\) on \(A_2 B_3\), etc. Determine the value of \(Z\) on each \(A_i B_j\) and determine the corresponding \(P(A_i B_j)\). From this, determine the distribution for \(Z\).
- Answer
-
A = [2 3 5]; B = [1 2 3]; a = rowcopy(A,3); b = colcopy(B,3); Z =a + b % Possible values of sum Z = X + Y Z = 3 4 6 4 5 7 5 6 8 PA = [0.3 0.6 0.1]; PB = [0.2 0.6 0.2]; pa= rowcopy(PA,3); pb = colcopy(PB,3); P = pa.*pb % Probabilities for various values P = 0.0600 0.1200 0.0200 0.1800 0.3600 0.0600 0.0600 0.1200 0.0200 [Z,PZ] = csort(Z,P); disp([Z;PZ]') % Distribution for Z = X + Y 3.0000 0.0600 4.0000 0.3000 5.0000 0.4200 6.0000 0.1400 7.0000 0.0600 8.0000 0.0200
Exercise \(\PageIndex{8}\)
For the random variables in Exercise, let \(W = XY\). Determine the value of \(W\) on each \(A_i B_j\) and determine the distribution of \(W\).
- Answer
-
XY = a.*b XY = 2 3 5 % XY values 4 6 10 6 9 15 W PW % Distribution for W = XY 2.0000 0.0600 3.0000 0.1200 4.0000 0.1800 5.0000 0.0200 6.0000 0.4200 9.0000 0.1200 10.0000 0.0600 15.0000 0.0200
Exercise \(\PageIndex{9}\)
A pair of dice is rolled.
- Let \(X\) be the minimum of the two numbers which turn up. Determine the distribution for \(X\)
- Let \(Y\) be the maximum of the two numbers. Determine the distribution for \(Y\).
- Let \(Z\) be the sum of the two numbers. Determine the distribution for \(Z\).
- Let \(W\) be the absolute value of the difference. Determine its distribution.
- Answer
-
t = 1:6; c = ones(6,6); [x,y] = meshgrid(t,t) x = 1 2 3 4 5 6 % x-values in each position 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 y = 1 1 1 1 1 1 % y-values in each position 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 m = min(x,y); % min in each position M = max(x,y); % max in each position s = x + y; % sum x+y in each position d = abs(x - y); % |x - y| in each position [X,fX] = csort(m,c) % sorts values and counts occurrences X = 1 2 3 4 5 6 fX = 11 9 7 5 3 1 % PX = fX/36 [Y,fY] = csort(M,c) Y = 1 2 3 4 5 6 fY = 1 3 5 7 9 11 % PY = fY/36 [Z,fZ] = csort(s,c) Z = 2 3 4 5 6 7 8 9 10 11 12 fZ = 1 2 3 4 5 6 5 4 3 2 1 %PZ = fZ/36 [W,fW] = csort(d,c) W = 0 1 2 3 4 5 fW = 6 10 8 6 4 2 % PW = fW/36
Exercise \(\PageIndex{10}\)
Minterm probabilities \(p(0)\) through \(p(15)\) for the class \(\{A, B , C, D\}\) are, in order,
0.072 0.048 0.018 0.012 0.168 0.112 0.042 0.028 0.062 0.048 0.028 0.010 0.170 0.110 0.040 0.
Determine the distribution for random variable
\(X = -5.3I_A - 2.5 I_B + 2.3 I_C + 4.2 I_D - 3.7\)
- Answer
-
% file npr06_10.m % Data for Exercise 6.2.10. pm = [ 0.072 0.048 0.018 0.012 0.168 0.112 0.042 0.028 ... 0.062 0.048 0.028 0.010 0.170 0.110 0.040 0.032]; c = [-5.3 -2.5 2.3 4.2 -3.7]; disp('Minterm probabilities are in pm, coefficients in c') npr06_10 Minterm probabilities are in pm, coefficients in c canonic Enter row vector of coefficients c Enter row vector of minterm probabilities pm Use row matrices X and PX for calculations Call for XDBN to view the distribution XDBN XDBN = -11.5000 0.1700 -9.2000 0.0400 -9.0000 0.0620 -7.3000 0.1100 -6.7000 0.0280 -6.2000 0.1680 -5.0000 0.0320 -4.8000 0.0480 -3.9000 0.0420 -3.7000 0.0720 -2.5000 0.0100 -2.0000 0.1120 -1.4000 0.0180 0.3000 0.0280 0.5000 0.0480 2.8000 0.0120
Exercise \(\PageIndex{11}\)
On a Tuesday evening, the Houston Rockets, the Orlando Magic, and the Chicago Bulls all have games (but not with one another). Let A be the event the Rockets win, \(B\) be the event the Magic win, and \(C\) be the event the Bulls win. Suppose the class {\(A, B, C\)} is independent, with respective probabilities 0.75, 0.70 0.8.  He wants to bet on the games. She decides to take him up on his bets as follows:
- $10 to 5 on the Rockets --- i.e. She loses five if the Rockets win and gains ten if they lose
- $10 to 5 against the Magic
- even $5 to 5 on the Bulls.
Ellen's winning may be expressed as the random variable
\(X = -5 I_A + 10 I_{A^c} + 10 I_B - 5 I_{B^c} - 5 I_C + 5I_{C^c} = -15 I_A + 15 I_B - 10 I_C + 10\)
Determine the distribution for \(X\). What are the probabilities Ellen loses money, breaks even, or comes out ahead?
- Answer
-
P = 0.01*[75 70 80]; c = [-15 15 -10 10]; canonic Enter row vector of coefficients c Enter row vector of minterm probabilities minprob(P) Use row matrices X and PX for calculations Call for XDBN to view the distribution disp(XDBN) -15.0000 0.1800 -5.0000 0.0450 0 0.4800 10.0000 0.1200 15.0000 0.1400 25.0000 0.0350 PXneg = (X<0)*PX' PXneg = 0.2250 PX0 = (X==0)*PX' PX0 = 0.4800 PXpos = (X>0)*PX' PXpos = 0.2950
Exercise \(\PageIndex{12}\)
The class {\(A, B, C, D\)} has minterm probabilities
\(pm = 0.001 *\) [5 7 6 8 9 14 22 33 21 32 50 75 86 129 201 302]
- Determine whether or not the class is independent.
- The random variable \(X = I_A + I_B + I_C + I_D\)counts the number of the events which occur on a trial. Find the distribution for X and determine the probability that two or more occur on a trial. Find the probability that one or three of these occur on a trial.
- Answer
-
npr06_12 Minterm probabilities in pm, coefficients in c a = imintest(pm) The class is NOT independent Minterms for which the product rule fails a = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 canonic Enter row vector of coefficients c Enter row vector of minterm probabilities pm Use row matrices X and PX for calculations Call for XDBN to view the distribution XDBN = 0 0.0050 1.0000 0.0430 2.0000 0.2120 3.0000 0.4380 4.0000 0.3020 P2 = (X>=2)*PX' P2 = 0.9520 P13 = ((X==1)|(X==3))*PX' P13 = 0.4810
Exercise \(\PageIndex{13}\)
James is expecting three checks in the mail, for $20, $26, and $33 dollars. Their arrivals are the events \(A, B, C\). Assume the class is independent, with respective probabilities 0.90, 0.75, 0.80. Then
\(X = 20 I_A + 26 I_B + 33 I_C\)
represents the total amount received. Determine the distribution for \(X\). What is the probability he receives at least $50? Less than $30?
- Answer
-
c = [20 26 33 0]; P = 0.01*[90 75 80]; canonic Enter row vector of coefficients c Enter row vector of minterm probabilities minprob(P) Use row matrices X and PX for calculations Call for XDBN to view the distribution disp(XDBN) 0 0.0050 20.0000 0.0450 26.0000 0.0150 33.0000 0.0200 46.0000 0.1350 53.0000 0.1800 59.0000 0.0600 79.0000 0.5400 P50 = (X>=50)*PX' P50 = 0.7800 P30 = (X <30)*PX' P30 = 0.0650
Exercise \(\PageIndex{14}\)
A gambler places three bets. He puts down two dollars for each bet. He picks up three dollars (his original bet plus one dollar) if he wins the first bet, four dollars if he wins the second bet, and six dollars if he wins the third. His net winning can be represented by the random variable
\(X = 3I_A + 4I_B + 6I_C - 6\), with \(P(A) = 0.5\), \(P(B) = 0.4\), \(P(C) = 0.3\)
Assume the results of the games are independent. Determine the distribution for \(X\).
- Answer
-
c = [3 4 6 -6]; P = 0.1*[5 4 3]; canonic Enter row vector of coefficients c Enter row vector of minterm probabilities minprob(P) Use row matrices X and PX for calculations Call for XDBN to view the distribution dsp(XDBN) -6.0000 0.2100 -3.0000 0.2100 -2.0000 0.1400 0 0.0900 1.0000 0.1400 3.0000 0.0900 4.0000 0.0600 7.0000 0.0600
Exercise \(\PageIndex{15}\)
Henry goes to a hardware store. He considers a power drill at $35, a socket wrench set at $56, a set of screwdrivers at $18, a vise at $24, and hammer at $8. He decides independently on the purchases of the individual items, with respective probabilities 0.5, 0.6, 0.7, 0.4, 0.9. Let \(X\) be the amount of his total purchases. Determine the distribution for \(X\).
- Answer
-
c = [35 56 18 24 8 0]; P = 0.1*[5 6 7 4 9]; canonic Enter row vector of coefficients c Enter row vector of minterm probabilities minprob(P) Use row matrices X and PX for calculations Call for XDBN to view the distribution disp(XDBN) 0 0.0036 8.0000 0.0324 18.0000 0.0084 24.0000 0.0024 26.0000 0.0756 32.0000 0.0216 35.0000 0.0036 42.0000 0.0056 43.0000 0.0324 50.0000 0.0504 53.0000 0.0084 56.0000 0.0054 59.0000 0.0024 61.0000 0.0756 64.0000 0.0486 67.0000 0.0216 74.0000 0.0126 77.0000 0.0056 80.0000 0.0036 82.0000 0.1134 85.0000 0.0504 88.0000 0.0324 91.0000 0.0054 98.0000 0.0084 99.0000 0.0486 106.0000 0.0756 109.0000 0.0126 115.0000 0.0036 117.0000 0.1134 123.0000 0.0324 133.0000 0.0084 141.0000 0.0756
Exercise \(\PageIndex{16}\)
A sequence of trials (not necessarily independent) is performed. Let \(E_i\) be the event of success on the \(i\)th component trial. We associate with each trial a "payoff function" \(X_i = aI_{E_i} + b I_{E_i^c}\). Thus, an amount \(a\) is earned if there is a success on the trial and an amount \(b\) (usually negative) if there is a failure. Let \(S_n\) be the number of successes in the \(n\) trials and \(W\) be the net payoff. Show that \(W = (a - b) S_n + bn\).
- Answer
-
\(X_i = aI_{E_i} + b(1 - I_{E_i}) = (a - b) I_{E_i} + b\)
\(W = \sum_{i = 1}^{n} X_i = (a - b) \sum_{i = 1}^{n} I_{E_i} + bn = (a - b) S_n + bn\)
Exercise \(\PageIndex{17}\)
A marker is placed at a reference position on a line (taken to be the origin); a coin is tossed repeatedly. If a head turns up, the marker is moved one unit to the right; if a tail turns up, the marker is moved one unit to the left.
-
Show that the position at the end of ten tosses is given by the random variable
\(X = \sum_{i = 1}^{10} I_{E_i} - \sum_{i = 1}^{10} I_{E_i^c} = 2 \sum_{i = 1}^{10} I_{E_i} - 10 = 2S_{10} - 10\)
where \(E_i\) is the event of a head on the \(i\)th toss and \(S_{10}\) is the number of heads in ten trials.
- After ten tosses, what are the possible positions and the probabilities of being in each?
- Answer
-
\(X_i = I_{E_i} - I_{E_i^c} = I_{E_i} - (1 - I_{E_i}) = 2I_{E_i} - 1\)
\(X = \sum_{i = 1}^{10} X_i = 2\sum_{i = 1}^{n} I_{E_i} - 10\)
S = 0:10; PS = ibinom(10,0.5,0:10); X = 2*S - 10; disp([X;PS]') -10.0000 0.0010 -8.0000 0.0098 -6.0000 0.0439 -4.0000 0.1172 -2.0000 0.2051 0 0.2461 2.0000 0.2051 4.0000 0.1172 6.0000 0.0439 8.0000 0.0098 10.0000 0.0010
Exercise \(\PageIndex{18}\)
Margaret considers five purchases in the amounts 5, 17, 21, 8, 15 dollars with respective probabilities 0.37, 0.22, 0.38, 0.81, 0.63. Anne contemplates six purchases in the amounts 8, 15, 12, 18, 15, 12 dollars, with respective probabilities 0.77, 0.52, 0.23, 0.41, 0.83, 0.58. Assume that all eleven possible purchases form an independent class.
- Determine the distribution for \(X\), the amount purchased by Margaret.
- Determine the distribution for \(Y\), the amount purchased by Anne.
- Determine the distribution for \(Z = X + Y\), the total amount the two purchase.
Suggestion for part (c). Let MATLAB perform the calculations.
- Answer
-
[r,s] = ndgrid(X,Y); [t,u] = ndgrid(PX,PY); z = r + s; pz = t.*u; [Z,PZ] = csort(z,pz);
% file npr06_18.m cx = [5 17 21 8 15 0]; cy = [8 15 12 18 15 12 0]; pmx = minprob(0.01*[37 22 38 81 63]); pmy = minprob(0.01*[77 52 23 41 83 58]); npr06_18 [X,PX] = canonicf(cx,pmx); [Y,PY] = canonicf(cy,pmy); [r,s] = ndgrid(X,Y); [t,u] = ndgrid(PX,PY); z = r + s; pz = t.*u; [Z,PZ] = csort(z,pz); a = length(Z) a = 125 % 125 different values plot(Z,cumsum(PZ)) % See figure Plotting details omitted


15.24B: Bacterial Infections of the Heart
- Recognize the causes and treatments for endocarditis
In a healthy individual, a bacteremia (where bacteria get into the blood stream through a minor cut or wound) would normally be cleared quickly with no adverse consequences. If a heart valve is damaged and covered with a piece of blood clot, the valve provides a place for the bacteria to attach themselves and an infection can be established. Endocarditis, or inflammation of the inner tissue of the heart, occurs as a result. The valves of the heart do not receive any dedicated blood supply. As a result, defensive immune mechanisms (such as white blood cells) cannot directly reach the valves via the bloodstream. When bacteria attaches to a valve surface and forms a vegetation, the host immune response is blunted. The lack of blood supply to the valves also has implications for treatment, since drugs also have difficulty reaching the infected valve. Normally, blood flows smoothly through these valves. If they have been damaged – from rheumatic fever, for example – the risk of bacterial attachment is increased.
Bacteremia caused by dental procedures (in most cases due to streptococci viridans, which reside in oral cavity), such as a cleaning or extraction of a tooth and from procedures involving the gastrointestinal or urinary tract can cause bacterial endocarditis. Intravenous drug abuse may also cause bacterial endocarditis from the aseptic introduction of skin bacteria.
Symptoms and signs of endocarditis vary, but prolonged fever (more then 2-3 days) without an obvious cause is a most important sign and should always be investigated in a child with congenital heart disease. Other signs and symptoms include poor appetite, feeling weak or tired, joint pains, skin rashes, and changes in the nature of a previously present heart murmur. The chance that these signs and symptoms are caused by endocarditis is more likely if they occur soon after a dental cleaning or procedure involving the gastrointestinal or urinary tract.
High dose antibiotics are administered by the intravenous route to maximize diffusion of antibiotic molecules into vegetation(s) from the blood filling the chambers of the heart. This is necessary because neither the heart valves nor the vegetations adherent to them are supplied by blood vessels. Antibiotics are continued for a long time, typically two to six weeks depending on the characteristics of the infection and the causative microorganisms.
Key Points
- Endocarditis occurs when bacteria grow on the edges of a heart defect or on the surface of an abnormal valve after the bacteria enter the blood stream, most commonly from dental procedures but also from procedures involving the gastrointestinal or urinary tract.
- The most important diagnostic test for endocarditis involves a positive blood culture. A blood culture is a small sample of blood drawn from the vein which is grown in a special solution so that bacteria can be detected.
- Symptoms and signs of endocarditis vary but include prolonged fever poor appetite, feeling weak or tired, joint pains, skin rashes, and changes in the nature of a previously present heart murmur.
- Treatment of bacterial endocarditis consists of a period of intravenous doses of appropriate antibiotics determined from blood tests under the supervision of an infectious disease specialist and cardiologist.
Key Terms
- Endocarditis : An inflammation of the interior lining of the heart or the endocardium and possibly the heart valves (pathology, cardiology).
- bacteremia : The medical condition of having bacteria in the bloodstream.


18.1: Bacterial Infections of the Skin and Eyes
- Identify the most common bacterial pathogens that cause infections of the skin and eyes
- Compare the major characteristics of specific bacterial diseases affecting the skin and eyes
Despite the skin’s protective functions, infections are common. Gram-positive Staphylococcus spp. are responsible for many of the most common skin infections. However, many skin conditions are not strictly associated with a single pathogen. Opportunistic pathogens of many types may infect skin wounds, and individual cases with identical symptoms may result from different pathogens or combinations of pathogens.
Staphylococcal Infections of the Skin
Key Concepts and Summary
- Staphylococcus causes many different types of skin infections, many of which occur when bacteria breach the skin barrier through a cut or wound.
- S. aureus are frequently associated with purulent skin infections that manifest as folliculitis , furuncles , or carbuncles . S. aureus is also a leading cause of staphylococcal scalded skin syndrome (SSSS).
- S. aureus is generally drug resistant and current MRSA strains are resistant to a wide range of antibiotics.
- Community-acquired and hospital-acquired staphyloccocal infections are an ongoing problem because many people are asymptomatic carriers.


8.1: Introduction
Health is influenced by a myriad of factors, many of them beyond genes or biology. Factors such as income and wealth, housing quality, access to greenspace, access to healthcare, education, stressful environments, and more - are often lumped into a category we call the Social Determinants of Health (SDOH). The U.S Department of Health and Human Services provides a broad definition as:
Social determinants of health (SDOH) are the conditions in the environments where people are born, live, learn, work, play, worship, and age that affect a wide range of health, functioning, and quality-of-life outcomes and risks.
Variance in these environments and conditions that people are born into and live in cause stark differences in health outcomes. Life expectancy overall is influenced by zip code. Life expectancy at birth can vary by 10 years or more between different census tracts within the same county (NCHS, 2022). Indeed, as the Vice President of the California Endowment (a non-profit health equity advocacy group) Dr. Tony Iton states: “When it comes to health, your zip code matters more than your genetic code.” (Iton, 2021).
In this chapter we will cover health disparities and inequities, each of the different categories of SDOH, and how these determinants affect health outcomes. Although the focus of this chapter will be to cover the SDOH in the United States of America, it is important to remember that health inequities stemming from SDOH span across the globe between and within different nations and continents - each with unique and yet interdependent causes.


12.E: Electromagnetism (Exercises)
1 . A particle with a charge of 1.0 C and a mass of 1.0 kg is observed moving past point P with a velocity \((1.0\ \text{m}/\text{s})\hat{\mathbf{x}}\). The electric field at point P is \((1.0\ \text{V}/\text{m})\hat{\mathbf{y}}\), and the magnetic field is \((2.0\ \text{T})\hat{\mathbf{y}}\). Find the force experienced by the particle.(answer check available at lightandmatter.com)
2 . For a positively charged particle moving through a magnetic field, the directions of the \(\mathbf{v}\), \(\mathbf{B}\), and \(\mathbf{F}\) vectors are related by a right-hand rule:
Make a three-dimensional model of the three vectors using pencils or rolled-up pieces of paper to represent the vectors assembled with their tails together. Make all three vectors perpendicular to each other. Now write down every possible way in which the rule could be rewritten by scrambling up the three symbols \(\mathbf{v}\), \(\mathbf{B}\), and \(\mathbf{F}\). Referring to your model, which are correct and which are incorrect?
3 . A charged particle is released from rest. We see it start to move, and as it gets going, we notice that its path starts to curve. Can we tell whether this region of space has \(\mathbf{E}\neq 0\), or \(\mathbf{B}\neq 0\), or both? Assume that no other forces are present besides the possible electrical and magnetic ones, and that the fields, if they are present, are uniform.
4 . A charged particle is in a region of space in which there is a uniform magnetic field \(\mathbf{B}=B\hat{\mathbf{z}}\). There is no electric field, and no other forces act on the particle. In each case, describe the future motion of the particle, given its initial velocity.
- \(\mathbf{v}_\text{o}=0\)
- \(\mathbf{v}_\text{o}=(1\ \text{m}/\text{s})\hat{\mathbf{z}}\)
- \(\mathbf{v}_\text{o}=(1\ \text{m}/\text{s})\hat{\mathbf{y}}\)
5
. (a) A line charge, with charge per unit length \(\lambda\), moves at velocity \(v\) along its own length. How much charge passes a given point in time \(dt\)? What is the resulting current? \hwans{hwans:linechargecurrent}
(b) Show that the units of your answer in part a work out correctly.
6
. Two parallel wires of length \(L\) carry currents \(I_1\) and \(I_2\). They are separated by a distance \(R\), and we assume \(R\) is much less than \(L\), so that our results for long, straight wires are accurate. The goal of this problem is to compute the magnetic forces acting between the wires.
(a) Neither wire can make a force on
itself
. Therefore, our first step in computing wire 1's force on wire 2 is to find the magnetic field made only by wire 1, in the space
occupied
by wire 2. Express this field in terms of the given quantities.(answer check available at lightandmatter.com)
(b) Let's model the current in wire 2 by pretending that there is a line charge inside it, possessing density per unit length \(\lambda_2\) and moving at velocity \(v_2\). Relate \(\lambda_2\) and \(v_2\) to the current \(I_2\), using the result of problem
5
a. Now find the magnetic force wire 1 makes on wire 2, in terms of \(I_1\), \(I_2\), \(L\), and \(R\). \hwans{hwans:forcebetweentwowires}
(c) Show that the units of the answer to part b work out to be newtons.
7 . Suppose a charged particle is moving through a region of space in which there is an electric field perpendicular to its velocity vector, and also a magnetic field perpendicular to both the particle's velocity vector and the electric field. Show that there will be one particular velocity at which the particle can be moving that results in a total force of zero on it. Relate this velocity to the magnitudes of the electric and magnetic fields. (Such an arrangement, called a velocity filter, is one way of determining the speed of an unknown particle.)
8 . The following data give the results of two experiments in which charged particles were released from the same point in space, and the forces on them were measured:
| q 1 = 1 µ C | \(\mathbf{v}_1=(1 m/s)\mathbf{\hat x}\) | \(\mathbf{F}_1=(-1 mN)\mathbf{\hat y}\) |
| q 2 = - 2 µ C | \(\mathbf{v}_2=(-1 m/s)\mathbf{\hat x}\) | \(\mathbf{F}_2=(-2 mN)\mathbf{\hat y}\) |
The data are insufficient to determine the magnetic field vector; demonstrate this by giving two different magnetic field vectors, both of which are consistent with the data.
9 . The following data give the results of two experiments in which charged particles were released from the same point in space, and the forces on them were measured:
| q 1 = 1 nC | \(\mathbf{v}_1=(1 m/s)\mathbf{\hat z}\) | \(\mathbf{F}_1=(5 mN)\mathbf{\hat x}+(2 mN)\mathbf{\hat y}\) |
| q 2 = 1 nC | \(\mathbf{v}_2=(3 m/s)\mathbf{\hat z}\) | \(\mathbf{F}_2=(10 mN)\mathbf{\hat x}+(4 mN)\mathbf{\hat y}\) |
Is there a nonzero electric field at this point? A nonzero magnetic field?
10 . This problem is a continuation of problem 6 . Note that the answer to problem 6 b is given on page 930.
- Interchanging the 1's and 2's in the answer to problem 6 b, what is the magnitude of the magnetic force from wire 2 acting on wire 1? Is this consistent with Newton's third law?
- Suppose the currents are in the same direction. Make a sketch, and use the right-hand rule to determine whether wire 1 pulls wire 2 towards it, or pushes it away.
- Apply the right-hand rule again to find the direction of wire 2's force on wire 1. Does this agree with Newton's third law?
- What would happen if wire 1's current was in the opposite direction compared to wire 2's?
11
. (a) In the photo of the vacuum tube apparatus in figure
o
on page 656, infer the direction of the magnetic field from the motion of the electron beam. (The answer is given in the answer to the self-check on that page.)
(b) Based on your answer to part a, find the direction of the currents in the coils.
(c) What direction are the electrons in the coils going?
(d) Are the currents in the coils repelling the currents consisting of the beam inside the tube, or attracting them? Check your answer by comparing with the result of problem
10
.
12 . A charged particle of mass \(m\) and charge \(q\) moves in a circle due to a uniform magnetic field of magnitude \(B\), which points perpendicular to the plane of the circle.
- Assume the particle is positively charged. Make a sketch showing the direction of motion and the direction of the field, and show that the resulting force is in the right direction to produce circular motion.
- Find the radius, \(r\), of the circle, in terms of \(m\), \(q\), \(v\), and \(B\).(answer check available at lightandmatter.com)
- Show that your result from part b has the right units.
- Discuss all four variables occurring on the right-hand side of your answer from part b. Do they make sense? For instance, what should happen to the radius when the magnetic field is made stronger? Does your equation behave this way?
- Restate your result so that it gives the particle's angular frequency, \(\omega\), in terms of the other variables, and show that \(v\) drops out.(answer check available at lightandmatter.com)
13 . Each figure represents the motion of a positively charged particle. The dots give the particles' positions at equal time intervals. In each case, determine whether the motion was caused by an electric force, a magnetic force, or a frictional force, and explain your reasoning. If possible, determine the direction of the magnetic or electric field. All fields are uniform. In (a), the particle stops for an instant at the upper right, but then comes back down and to the left, retracing the same dots. In (b), it stops on the upper right and stays there.
14 . One model of the hydrogen atom has the electron circling around the proton at a speed of \(2.2\times10^6\) m/s, in an orbit with a radius of 0.05 nm. (Although the electron and proton really orbit around their common center of mass, the center of mass is very close to the proton, since it is 2000 times more massive. For this problem, assume the proton is stationary.)
- Treat the circling electron as a current loop, and calculate the current.
- Estimate the magnetic field created at the center of the atom by the electron.(answer check available at lightandmatter.com)
- Does the proton experience a nonzero force from the electron's magnetic field? Explain.
- Does the electron experience a magnetic field from the proton? Explain.
- Does the electron experience a magnetic field created by its own current? Explain.
- Is there an electric force acting between the proton and electron? If so, calculate it.(answer check available at lightandmatter.com)
- Is there a gravitational force acting between the proton and electron? If so, calculate it.
- An inward force is required to keep the electron in its orbit -- otherwise it would obey Newton's first law and go straight, leaving the atom. Based on your answers to the previous parts, which force or forces (electric, magnetic and gravitational) contributes significantly to this inward force? (Based on a problem by Arnold Arons.)
15 . The equation \(B_z=\beta kIA/c^2r^3\) was found on page 666 for the distant field of a dipole. Show, as asserted there, that the constant \(\beta\) must be unitless.
16 . The following data give the results of three experiments in which charged particles were released from the same point in space, and the forces on them were measured:
| q 1 = 1 C | \(\mathbf{v}_1=0\) | \(\mathbf{F}_1=(1 N)\mathbf{\hat y}\) |
| q 2 =1 C | \(\mathbf{v}_2=(1 m/s)\mathbf{\hat x}\) | \(\mathbf{F}_2=(1 N)\mathbf{\hat y}\) |
| q 3 =1 C | \(\mathbf{v}_3=(1 m/s)\mathbf{\hat z}\) | \(\mathbf{F}_3=0\) |
Determine the electric and magnetic fields.(answer check available at lightandmatter.com)
17 . If you put four times more current through a solenoid, how many times more energy is stored in its magnetic field?(answer check available at lightandmatter.com)
18
. A Helmholtz coil is defined as a pair of identical circular coils lying in parallel planes and separated by a distance, \(h\), equal to their radius, \(b\). (Each coil may have more than one turn of wire.) Current circulates in the same direction in each coil, so the fields tend to reinforce each other in the interior region. This configuration has the advantage of being fairly open, so that other apparatus can be easily placed inside and subjected to the field while remaining visible from the outside. The choice of \(h=b\) results in the most uniform possible field near the center. A photograph of a Helmholtz coil was shown in figure
o
on page 656.
(a) Find the percentage drop in the field at the center of one coil, compared to the full strength at the center of the whole apparatus. (answer check available at lightandmatter.com)
(b) What value of \(h\) (not equal to \(b)\) would make this percentage difference equal to zero?(answer check available at lightandmatter.com)
19 . The figure shows a nested pair of circular wire loops used to create magnetic fields. (The twisting of the leads is a practical trick for reducing the magnetic fields they contribute, so the fields are very nearly what we would expect for an ideal circular current loop.) The coordinate system below is to make it easier to discuss directions in space. One loop is in the \(y-z\) plane, the other in the \(x-y\) plane. Each of the loops has a radius of 1.0 cm, and carries 1.0 A in the direction indicated by the arrow.
- Calculate the magnetic field that would be produced by one such loop, at its center. (answer check available at lightandmatter.com)
- Describe the direction of the magnetic field that would be produced, at its center, by the loop in the \(x-y\) plane alone.
- Do the same for the other loop.
- Calculate the magnitude of the magnetic field produced by the two loops in combination, at their common center. Describe its direction.(answer check available at lightandmatter.com)
20 . Four long wires are arranged, as shown, so that their cross-section forms a square, with connections at the ends so that current flows through all four before exiting. Note that the current is to the right in the two back wires, but to the left in the front wires. If the dimensions of the cross-sectional square (height and front-to-back) are \(b\), find the magnetic field (magnitude and direction) along the long central axis.(answer check available at lightandmatter.com)
21 . In problem 16 , the three experiments gave enough information to determine both fields. Is it possible to design a procedure so that, using only two such experiments, we can always find \(\mathbf{E}\) and \(\mathbf{B}\)? If so, design it. If not, why not?
22 . Use the Biot-Savart law to derive the magnetic field of a long, straight wire, and show that this reproduces the result of example 6 on page 658.
23
. (a) Modify the calculation on page 663 to determine the component of the magnetic field of a sheet of charge that is perpendicular to the sheet.(answer check available at lightandmatter.com)
(b) Show that your answer has the right units.
(c) Show that your answer approaches zero as \(z\) approaches infinity.
(d) What happens to your answer in the case of \(a=b\)? Explain why this makes sense.
24 . Consider two solenoids, one of which is smaller so that it can be put inside the other. Assume they are long enough so that each one only contributes significantly to the field inside itself, and the interior fields are nearly uniform. Consider the configuration where the small one is inside the big one with their currents circulating in the same direction, and a second configuration in which the currents circulate in opposite directions. Compare the energies of these configurations with the energy when the solenoids are far apart. Based on this reasoning, which configuration is stable, and in which configuration will the little solenoid tend to get twisted around or spit out? \hwhint{hwhint:nestedsolenoids}
25
. (a) A solenoid can be imagined as a series of circular current loops that are spaced along their common axis. Integrate the result of example
12
on page 671 to show that the field on the axis of a solenoid can be written as \(B=(2\pi k\eta/c^2)(\cos\beta+\cos\gamma)\), where the angles \(\beta\) and \(\gamma\) are defined in the figure.
(b) Show that in the limit where the solenoid is very long, this exact result agrees with the approximate one derived in example
13
on page 674 using Ampère's law.
(c) Note that, unlike the calculation using Ampère's law, this one is valid at points that are near the mouths of the solenoid, or even outside it entirely. If the solenoid is long, at what point on the axis is the field equal to one half of its value at the center of the solenoid?
(d) What happens to your result when you apply it to points that are very far away from the solenoid? Does this make sense?
26 . The first step in the proof of Ampère's law on page 675 is to show that Ampère's law holds in the case shown in figure f /1, where a circular Ampèrian loop is centered on a long, straight wire that is perpendicular to the plane of the loop. Carry out this calculation, using the result for the field of a wire that was established without using Ampère's law.
27 . A certain region of space has a magnetic field given by \(\mathbf{B}=bx\hat{\mathbf{y}}\). Find the electric current flowing through the square defined by \(z=0\), \(0\le x \le a\), and \(0\le y \le a\).(answer check available at lightandmatter.com)
28 . Perform a calculation similar to the one in problem 54 , but for a logarithmic spiral, defined by \(r=we^{u\theta}\), and show that the field is \(B=(kI/c^2u)(1/a-1/b)\). Note that the solution to problem 54 is given in the back of the book.
29
. (a) For the geometry described in example
8
on page 661, find the field at a point the lies in the plane of the wires, but not between the wires, at a distance \(b\) from the center line. Use the same technique as in that example.
(b) Now redo the calculation using the technique demonstrated on page 666. The integrals are nearly the same, but now the reasoning is reversed: you already know \(\beta=1\), and you want to find an unknown field. The only difference in the integrals is that you are tiling a different region of the plane in order to mock up the currents in the two wires. Note that you can't tile a region that contains a point of interest, since the technique uses the field of a distant dipole.(answer check available at lightandmatter.com)
30
. (a) A long, skinny solenoid consists of \(N\) turns of wire wrapped uniformly around a hollow cylinder of length \(\ell\) and cross-sectional area \(A\). Find its inductance.(answer check available at lightandmatter.com)
(b) Show that your answer has the right units to be an inductance.
31
. Consider two solenoids, one of which is smaller so that it can be put inside the other. Assume they are long enough to act like ideal solenoids, so that each one only contributes significantly to the field inside itself, and the interior fields are nearly uniform. Consider the configuration where the small one is partly inside and partly hanging out of the big one, with their currents circulating in the same direction. Their axes are constrained to coincide.
(a) Find the difference in the magnetic energy between the configuration where the solenoids are separate and the configuration where the small one is inserted into the big one. Your equation will include the length \(x\) of the part of the small solenoid that is inside the big one, as well as other relevant variables describing the two solenoids.(answer check available at lightandmatter.com)
(b) Based on your answer to part a, find the force acting between the solenoids.(answer check available at lightandmatter.com)
32 . Verify Ampère's law in the case shown in the figure, assuming the known equation for the field of a wire. A wire carrying current \(I\) passes perpendicularly through the center of the rectangular Ampèrian surface. The length of the rectangle is infinite, so it's not necessary to compute the contributions of the ends.
33
. The purpose of this problem is to find how the gain of a transformer depends on its construction.
(a) The number of loops of wire, \(N\), in a solenoid is changed, while keeping the length constant. How does the impedance depend on \(N\)? State your answer as a proportionality, e.g., \(Z\propto N^3\) or \(Z\propto N^{-5}\).
(b) For a given AC voltage applied across the inductor, how does the magnetic field depend on \(N\)? You need to take into account both the dependence of a solenoid's field on \(N\) for a given current and your answer to part a, which affects the current.
(c) Now consider a transformer consisting of two solenoids. The input side has \(N_1\) loops, and the output \(N_2\). We wish to find how the output voltage \(V_2\) depends on \(N_1\), \(N_2\), and the input voltage \(V_1\). The text has already established \(V_2\propto V_1N_2\), so it only remains to find the dependence on \(N_1\). Use your result from part b to accomplish this. The ratio \(V_2/V_1\) is called the voltage gain.
34 . Problem 33 dealt with the dependence of a transformer's gain on the number of loops of wire in the input solenoid. Carry out a similar analysis of how the gain depends on the frequency at which the circuit is operated.
35
. A U-shaped wire makes electrical contact with a second, straight wire, which rolls along it to the right, as shown in the figure. The whole thing is immersed in a uniform magnetic field, which is perpendicular to the plane of the circuit. The resistance of the rolling wire is much greater than that of the U.
(a) Find the direction of the force on the wire based on conservation of energy.
(b) Verify the direction of the force using right-hand rules.
(c) Find magnitude of the force acting on the wire. There is more than one way to do this, but please do it using Faraday's law (which works even though it's the Ampèrian surface itself that is changing, rather than the field).(answer check available at lightandmatter.com)
(d) Consider how the answer to part a would have changed if the direction of the field had been reversed, and also do the case where the direction of the rolling wire's motion is reversed. Verify that this is in agreement with your answer to part c.
36 . A charged particle is in motion at speed \(v\), in a region of vacuum through which an electromagnetic wave is passing. In what direction should the particle be moving in order to minimize the total force acting on it? Consider both possibilities for the sign of the charge. (Based on a problem by David J. Raymond.)
37
. A wire loop of resistance \(R\) and area \(A\), lying in the \(y-z\) plane, falls through a nonuniform magnetic field \(\mathbf{B}=kz\hat{\mathbf{x}}\), where \(k\) is a constant. The \(z\) axis is vertical.
(a) Find the direction of the force on the wire based on conservation of energy.
(b) Verify the direction of the force using right-hand rules.
(c) Find the magnetic force on the wire.(answer check available at lightandmatter.com)
38 . A capacitor has parallel plates of area \(A\), separated by a distance \(h\). If there is a vacuum between the plates, then Gauss's law gives \(E=4\pi k\sigma=4\pi kq/A\) for the field between the plates, and combining this with \(E=V/h\), we find \(C=q/V=(1/4\pi k)A/h\). (a) Generalize this derivation to the case where there is a dielectric between the plates. (b) Suppose we have a list of possible materials we could choose as dielectrics, and we wish to construct a capacitor that will have the highest possible energy density, \(U_e/v\), where \(v\) is the volume. For each dielectric, we know its permittivity \(\epsilon\), and also the maximum electric field \(E\) it can sustain without breaking down and allowing sparks to cross between the plates. Write the maximum energy density in terms of these two variables, and determine a figure of merit that could be used to decide which material would be the best choice.
39
. (a) For each term appearing on the right side of Maxwell's equations, give an example of an everyday situation it describes.
(b) Most people doing calculations in the SI system of units don't use \(k\) and \(k/c^2\). Instead, they express everything in terms of the constants
Rewrite Maxwell's equations in terms of these constants, eliminating \(k\) and \(c\) everywhere.
40
. (a) Prove that in an electromagnetic plane wave, half the energy is in the electric field and half in the magnetic field.
(b) Based on your result from part a, find the proportionality constant in the relation \(d\mathbf{p}\propto\mathbf{E}\times\mathbf{B}dv\), where \(d \mathbf{p}\) is the momentum of the part of a plane light wave contained in the volume \(dv\). The vector \(\mathbf{E}\times\mathbf{B}\) is known as the Poynting vector. (To do this problem, you need to know the relativistic relationship between the energy and momentum of a beam of light.)(answer check available at lightandmatter.com)
41
. (a) A beam of light has cross-sectional area \(A\) and power \(P\), i.e., \(P\) is the number of joules per second that enter a window through which the beam passes. Find the energy density \(U/v\) in terms of \(P\), \(A\), and universal constants.
(b) Find \(\tilde{\mathbf{E}}\) and \(\tilde{\mathbf{B}}\), the amplitudes of the electric and magnetic fields, in terms of \(P\), \(A\), and universal constants (i.e., your answer should
not
include \(U\) or \(v\)). You will need the result of problem
40
a. A real beam of light usually consists of many short wavetrains, not one big sine wave, but don't worry about that.(answer check available at lightandmatter.com)\hwhint{hwhint:solarconstant}
(c) A beam of sunlight has an intensity of \(P/A=1.35\times10^3\ \text{W}/\text{m}^2\), assuming no clouds or atmospheric absorption. This is known as the solar constant. Compute \(\tilde{\mathbf{E}}\) and \(\tilde{\mathbf{B}}\), and compare with the strengths of static fields you experience in everyday life: \(E \sim 10^6\ \text{V}/\text{m}\) in a thunderstorm, and \(B \sim 10^{-3}\) T for the Earth's magnetic field.(answer check available at lightandmatter.com)
42 . The circular parallel-plate capacitor shown in the figure is being charged up over time, with the voltage difference across the plates varying as \(V=st\), where \(s\) is a constant. The plates have radius \(b\), and the distance between them is \(d\). We assume \(d \ll b\), so that the electric field between the plates is uniform, and parallel to the axis. Find the induced magnetic field at a point between the plates, at a distance \(R\) from the axis. \hwhint{hwhint:circularcap}(answer check available at lightandmatter.com)
43
. A positively charged particle is released from rest at the origin at \(t=0\), in a region of vacuum through which an electromagnetic wave is passing. The particle accelerates in response to the wave. In this region of space, the wave varies as \(\mathbf{E}=\hat{\mathbf{x}}\tilde{E}\sin\omega t\), \(\mathbf{B}=\hat{\mathbf{y}}\tilde{B}\sin\omega t\), and we assume that the particle has a relatively large value of \(m/q\), so that its response to the wave is sluggish, and it never ends up moving at any speed comparable to the speed of light. Therefore we don't have to worry about the spatial variation of the wave; we can just imagine that these are uniform fields imposed by some external mechanism on this region of space.
(a) Find the particle's coordinates as functions of time.(answer check available at lightandmatter.com)
(b) Show that the motion is confined to \(-z_{max}\leq z \leq z_{max}\), where \(z_{max} = 1.101\left(q^2\tilde{E}\tilde{B}/m^2\omega^3\right)\).
44 . Electromagnetic waves are supposed to have their electric and magnetic fields perpendicular to each other. (Throughout this problem, assume we're talking about waves traveling through a vacuum, and that there is only a single sine wave traveling in a single direction, not a superposition of sine waves passing through each other.) Suppose someone claims they can make an electromagnetic wave in which the electric and magnetic fields lie in the same plane. Prove that this is impossible based on Maxwell's equations.
45 . Repeat the self-check on page 710, but with one change in the procedure: after we charge the capacitor, we open the circuit, and then continue with the observations.
46 . On page 713, I proved that \(\mathbf{H}_{\parallel,1}=\mathbf{H}_{\parallel,2}\) at the boundary between two substances if there is no free current and the fields are static. In fact, each of Maxwell's four equations implies a constraint with a similar structure. Some are constraints on the field components parallel to the boundary, while others are constraints on the perpendicular parts. Since some of the fields referred to in Maxwell's equations are the electric and magnetic fields \(\mathbf{E}\) and \(\mathbf{B}\), while others are the auxiliary fields \(\mathbf{D}\) and \(\mathbf{H}\), some of the constraints deal with \(\mathbf{E}\) and \(\mathbf{B}\), others with \(\mathbf{D}\) and \(\mathbf{H}\). Find the other three constraints.
47 . (a) Figure j on page 714 shows a hollow sphere with \(\mu/\mu_\text{o}=x\), inner radius \(a\), and outer radius \(b\), which has been subjected to an external field \(\mathbf{B}_\text{o}\). Finding the fields on the exterior, in the shell, and on the interior requires finding a set of fields that satisfies five boundary conditions: (1) far from the sphere, the field must approach the constant \(\mathbf{B}_\text{o}\); (2) at the outer surface of the sphere, the field must have \(\mathbf{H}_{\parallel,1}=\mathbf{H}_{\parallel,2}\), as discussed on page 713; (3) the same constraint applies at the inner surface of the sphere; (4) and (5) there is an additional constraint on the fields at the inner and outer surfaces, as found in problem 46 . The goal of this problem is to find the solution for the fields, and from it, to prove that the interior field is uniform, and given by
This is a very difficult problem to solve from first principles, because it's not obvious what form the fields should have, and if you hadn't been told, you probably wouldn't have guessed that the interior field would be uniform. We could, however, guess that once the sphere becomes polarized by the external field, it would become a dipole, and at \(r\gg b\), the field would be a uniform field superimposed on the field of a dipole. It turns out that even close to the sphere, the solution has exactly this form. In order to complete the solution, we need to find the field in the shell (\(a\lt r\lt b\)), but the only way this field could match up with the detailed angular variation of the interior and exterior fields would be if it was also a superposition of a uniform field with a dipole field. The final result is that we have four unknowns: the strength of the dipole component of the external field, the strength of the uniform and dipole components of the field within the shell, and the strength of the uniform interior field. These four unknowns are to be determined by imposing constraints (2) through (5) above.
(b) Show that the expression from part a has physically reasonable behavior in its dependence on \(x\) and \(a/b\).
48
. Two long, parallel strips of thin metal foil form a configuration like a long, narrow sandwich. The air gap between them has height \(h\), the width of each strip is \(w\), and their length is \(\ell\). Each strip carries current \(I\), and we assume for concreteness that the currents are in opposite directions, so that the magnetic force, \(F\), between the strips is repulsive.
(a) Find the force in the limit of \(w\gg h\).(answer check available at lightandmatter.com)
(b) Find the force in the limit of \(w\ll h\), which is like two ordinary wires.
(c) Discuss the relationship between the two results.
49 . Suppose we are given a permanent magnet with a complicated, asymmetric shape. Describe how a series of measurements with a magnetic compass could be used to determine the strength and direction of its magnetic field at some point of interest. Assume that you are only able to see the direction to which the compass needle settles; you cannot measure the torque acting on it.
50
. On page 680, the curl of \(x\hat{\mathbf{y}}\) was computed. Now consider the fields \(x\hat{\mathbf{x}}\) and \(y\hat{\mathbf{y}}\).
(a) Sketch these fields.
(b) Using the same technique of explicitly constructing a small square, prove that their curls are both zero. Do not use the component form of the curl; this was one step in
deriving
the component form of the curl.
51
. If you watch a movie played backwards, some vectors reverse their direction. For instance, people walk backwards, with their velocity vectors flipped around. Other vectors, such as forces, keep the same direction, e.g., gravity still pulls down. An electric field is another example of a vector that doesn't turn around: positive charges are still positive in the time-reversed universe, so they still make diverging electric fields, and likewise for the converging fields around negative charges.
(a) How does the momentum of a material object behave under time-reversal?(solution in the pdf version of the book)
(b) The laws of physics are still valid in the time-reversed universe. For example, show that if two material objects are interacting, and momentum is conserved, then momentum is still conserved in the time-reversed universe.(solution in the pdf version of the book)
(c) Discuss how currents and magnetic fields would behave under time reversal. \hwhint{hwhint:timereversalem}
(d) Similarly, show that the equation \(d\mathbf{p}\propto\mathbf{E}\times\mathbf{B}\) is still valid under time reversal.
52
. This problem is a more advanced exploration of the time-reversal ideas introduced in problem
51
.
(a) In that problem, we assumed that charge did not flip its sign under time reversal. Suppose we make the opposite assumption, that charge
does
change its sign. This is an idea introduced by Richard Feynman: that antimatter is really matter traveling backward in time! Determine the time-reversal properties of \(\mathbf{E}\) and \(\mathbf{B}\) under this new assumption, and show that \(d\mathbf{p}\propto\mathbf{E}\times\mathbf{B}\) is still valid under time-reversal.
(b) Show that Maxwell's equations are time-reversal symmetric, i.e., that if the fields \(\mathbf{E}(x,y,z,t)\) and \(\mathbf{B}(x,y,z,t)\) satisfy Maxwell's equations, then so do \(\mathbf{E}(x,y,z,-t)\) and \(\mathbf{B}(x,y,z,-t)\). Demonstrate this under both possible assumptions about charge, \(q\rightarrow q\) and \(q\rightarrow -q\).
53
. The purpose of this problem is to prove that the constant of proportionality \(a\) in the equation \(dU_m=aB^2 dv\), for the energy density of the magnetic field, is given by \(a=c^2/8\pi k\) as asserted on page 665. The geometry we'll use consists of two sheets of current, like a sandwich with nothing in between but some vacuum in which there is a magnetic field. The currents are in opposite directions, and we can imagine them as being joined together at the ends to form a complete circuit, like a tube made of paper that has been squashed almost flat. The sheets have lengths \(L\) in the direction parallel to the current, and widths \(w\). They are separated by a distance \(d\), which, for convenience, we assume is small compared to \(L\) and \(w\). Thus each sheet's contribution to the field is uniform, and can be approximated by the expression \(2\pi k\eta/c^2\).
(a) Make a drawing similar to the one in figure
11.2.1
on page 664, and show that in this opposite-current configuration, the magnetic fields of the two sheets reinforce in the region between them, producing double the field, but cancel on the outside.
(b) By analogy with the case of a single strand of wire, one sheet's force on the other is \(ILB_1\), were \(I=\eta w\) is the total current in one sheet, and \(B_1=B/2\) is the field contributed by only one of the sheets, since the sheet can't make any net force on itself. Based on your drawing and the right-hand rule, show that this force is repulsive.
For the rest of the problem, consider a process in which the sheets start out touching, and are then separated to a distance \(d\). Since the force between the sheets is repulsive, they do mechanical work on the outside world as they are separated, in much the same way that the piston in an engine does work as the gases inside the cylinder expand. At the same time, however, there is an induced emf which would tend to extinguish the current, so in order to maintain a constant current, energy will have to be drained from a battery. There are three types of energy involved: the increase in the magnetic field energy, the increase in the energy of the outside world, and the decrease in energy as the battery is drained. (We assume the sheets have very little resistance, so there is no ohmic heating involved.)(answer check available at lightandmatter.com)
(c) Find the mechanical work done by the sheets, which equals the increase in the energy of the outside world. Show that your result can be stated in terms of \(\eta\), the final volume \(v=wLd\), and nothing else but numerical and physical constants.(answer check available at lightandmatter.com)
(d) The power supplied by the battery is \(P=I\Gamma_E\) (like \(P=I\Delta V\), but with an emf instead of a voltage difference), and the circulation is given by \(\Gamma=-d\Phi_B/dt\). The negative sign indicates that the battery is being drained. Calculate the energy supplied by the battery, and, as in part c, show that the result can be stated in terms of \(\eta\), \(v\), and universal constants.(answer check available at lightandmatter.com)
(e) Find the increase in the magnetic-field energy, in terms of \(\eta\), \(v\), and the unknown constant \(a\).(answer check available at lightandmatter.com)
(f) Use conservation of energy to relate your answers from parts c, d, and e, and solve for \(a\).(answer check available at lightandmatter.com)
54 . Magnet coils are often wrapped in multiple layers. The figure shows the special case where the layers are all confined to a single plane, forming a spiral. Since the thickness of the wires (plus their insulation) is fixed, the spiral that results is a mathematical type known as an Archimedean spiral, in which the turns are evenly spaced. The equation of the spiral is \(r=w\theta\), where \(w\) is a constant. For a spiral that starts from \(r=a\) and ends at \(r=b\), show that the field at the center is given by \((kI/c^2w)\ln b/a\).(solution in the pdf version of the book)
Contributors and Attributions
Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CC-BY-SA license.


5.5: The Negative Exponent Rule
In section 5.3, the exponent of the number in the numerator was greater than the exponent of the number in the denominator. In section 5.4, the exponent of the number in the numerator was equal to the exponent of the number in the denominator. In section 5.5, the exponent of the number in the denominator may be greater than the exponent of the number in the numerator.
For any non zero real number a and any integer n, the negative exponent rule is the following
\(a^{−n}= \dfrac{1 }{a^n} or \dfrac{1 }{a^{−n}} = a^n\)
It is poor form in mathematics to leave negative exponents in the answer. All answers will always be simplified to show positive exponents.
How does this work?
Recall:
\[\begin{align*} 2^3 &= 2 \cdot 2 \cdot 2 = 8 \\[4pt] 2^2 &= 2 \cdot 2 = 4 \\[4pt] 2^1 &= 2 = 2 \\[4pt] 2^0 &= 1 \end{align*} \nonumber \]
What happens with negative exponents?
\[\begin{align*} 2^{−1 } &= \dfrac{1 }{2^1 }= \dfrac{1 }{2} \\[4pt] 2^{−2 } &= \dfrac{1}{2^2} = \dfrac{1}{ 4} \end{align*} \nonumber \]
Recall: From the last section,
\[\begin{align*} x^3 = \textcolor{blue}{x \cdot x \cdot x} \\[4pt] x^5 &= \textcolor{red}{x \cdot x \cdot x \cdot x \cdot x} \end{align*} \nonumber \]
Their quotient:
\(\dfrac{x^3 }{x^5} = \dfrac{x \cdot x \cdot x }{x \cdot x \cdot x \cdot x \cdot x }= \dfrac{\textcolor{blue}{\cancel{x \cdot x\cdot x }}}{\textcolor{red}{\cancel{x \cdot x\cdot x \cdot x \cdot x }}}=\dfrac{ 1 }{\textcolor{red}{x \cdot x }}= \dfrac{1 }{x^2}\)
Apply the quotient rule to obtain an equivalent result.
\(\dfrac{x^3 }{x^5} = x^{3−5 }= x^{−2}\)
Using the negative exponent rule.
\(x^{−2 }= \dfrac{1 }{x^2}\).
Review the following examples to help understand the process of simplifying using the quotient rule of exponents and the negative exponent rule.
Hint: Be patient, take your time and be careful when simplifying!
Simplify the following expression to a single base with only positive exponents.
\(\dfrac{t^5}{ t^11}\)
Solution
\(t^{5−11 }= t^{−6 }= \dfrac{1 }{t^6}\)
Simplify the following expression to a single base with only positive exponents.
\(\dfrac{x^{3} \cdot x^{11} }{x \cdot x^{15}}\)
Solution
\(\dfrac{x^{ 3+11 }}{x^{1+15 }}= \dfrac{x^14 }{x^16 }= x^{14−16 }= x^{−2 }= \dfrac{1}{x^2}\)
Simplify the following expression to a single base with only positive exponents.
\(\dfrac{2y^3 }{7y^7}\)
Solution
\(\dfrac{2 }{7} \cdot \dfrac{y^3 }{y^7 }= \dfrac{2 }{7} \cdot y^{3−7 }= \dfrac{2 }{7 }\cdot y^{−4} = \dfrac{2 }{7 }\cdot \dfrac{1 }{y^4 }= \dfrac{2 }{7y^4}\)
Simplify the following expression to a single base with only positive exponents.
\(-\dfrac{\sqrt{3}z^6}{ z^7}\)
Solution
\(− \sqrt{3} \cdot \dfrac{z^6 }{z^7} = − \sqrt{3} \cdot z^{6−7 }= − \sqrt{3} \cdot z^{−1} = − \sqrt{3} \cdot \dfrac{1 }{z} = − \dfrac{\sqrt{3}}{z}\)
In Examples 3 and 4, factor out the constant to see the common bases clearly.
Simplify the following expression to a single base with only positive exponents.
\(\dfrac{1}{ a^{−9}}\)
Solution
\(a^9\)
Simplify the following expression to a single base with only positive exponents.
\(\dfrac{x^3 }{x^{−5}}\)
Solution
\(x^{3−(−5) }= x^{3+5 }= x^{8}\)
Simplify the following expression to a single base with only positive exponents.
\(\dfrac{c^{−7 }}{c^{−3}}\)
Solution
\(c^{(−7)−(−3) }= c^{−7+3 }= c ^{−4} = \dfrac{1 }{c^4}\)
In examples 6 and 7, the quotient rule of exponents was used before changing exponents to positive exponents. The same results are obtained by expanding and changing exponents to positive exponents first and then applying the quotient rule of exponents.
Simplify the following expressions to a single base with only positive exponents.
- \(\dfrac{p^4}{ p^{13}}\)
- \(-\dfrac{k^2 \cdot k^3 }{k^7 \cdot k^8}\)
- \(\dfrac{5(x + y)^3 }{2(x + y)^{13}}\)
- −\(\dfrac{\sqrt{8}y^3}{ y^{−3}}\)
- \(\dfrac{a^{−7}}{ a^2 \cdot a^{−5}}\)
- \(\dfrac{x^{−7}}{ x^5}\)


2.5: Properties of Exponents
In the exponential expression \(a^n\), the number \(a\) is called the base , while the number \(n\) is called the exponent .
Let \(a\) be any real number and let \(n\) be any whole number. If \(n \neq 0\), then:
\[a^{n}=\underbrace{a \cdot a \cdot a \cdots \cdot a}_{n \text { times }} \nonumber \]
That is, to calculate \(a^n\), write \(a\) as a factor \(n\) times. In the case where \(a \neq 0\), but \(n = 0\), then we define:
\[a^0 =1 \nonumber \]
For example, raising a number to the fifth power requires that we repeat the number as a factor five times (see Figure \(\PageIndex{1}\)).
\[\begin{aligned}(-2)^{5} &=(-2)(-2)(-2)(-2)(-2) \\[4pt] &=-32 \end{aligned} \nonumber \]
Raising a number to the fourth power requires that we repeat that number as a factor four times.
\[\begin{aligned}\left(-\dfrac{1}{2}\right)^{4} &=\left(-\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right) \\ &=\dfrac{1}{16} \end{aligned} \nonumber \]
Remember that if there is a negative in front of the term without parentheses that the negative is not raised to the power.
\[\begin{aligned}-2^4 &=-1 \cdot 2^4 \\&= -1 \cdot (2)(2)(2)(2) \\&= -16 \end{aligned} \nonumber \]
As a final example, note that \(10^0 = 1\), but \(0^0\) is undefined (see Figure \(\PageIndex{2}\)).
For those who may be wondering why \(a^0 = 1\), provided \(a \neq 0\) , here is a nice argument. First, note that \(a^1 = a\), so:
\[a \cdot a^{0}=a^{1} \cdot a^{0} \nonumber \]
On the right, repeat the base and add the exponents.
\[a \cdot a^{0}=a^{1} \nonumber \]
Or equivalently:
\[a \cdot a^{0}=a \nonumber \]
Now, divide both sides by \(a\), which is permissible if \(a \neq 0\) .
\[\dfrac{a \cdot a^{0}}{a}=\dfrac{a}{a} \nonumber \]
Simplify both sides:
\[a^0 =1 \nonumber \]
Multiplying With Like Bases
In the expression \(a^n\), the number \(a\) is called the base and the number \(n\) is called the exponent . Frequently, we’ll be required to multiply two exponential expressions with like bases, such as \(x^{3} \cdot x^{4}\). Recall that the exponent tells us how many times to write each base as a factor, so we can write:
\[\begin{aligned} x^{3} \cdot x^{4} &=(x \cdot x \cdot x) \cdot(x \cdot x \cdot x \cdot x) \\ &=x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \\ &=x^{7} \end{aligned} \nonumber \]
Note that we are simply counting the number of times that \(x\) occurs as a factor. First, we have three \(x’s\), then four \(x’s\), for a total of seven \(x’s\). However a little thought tells us that it is much quicker to simply add the exponents to reveal the total number of times \(x\) occurs as a factor.
\[\begin{aligned} x^{3} \cdot x^{4} &=x^{3+4} \\ &=x^{7} \end{aligned} \nonumber \]
The preceding discussion is an example of the following general property of exponents.
To multiply two exponential expressions with like bases, repeat the base and add the exponents.
\[a^{m} \cdot a^{n}=a^{m+n} \nonumber \]
Simplify each of the following expressions:
- \(y^{4} \cdot y^{8}\)
- \(2^{3} \cdot 2^{5}\)
- \((x+y)^{2}(x+y)^{7}\)
Solution
In each example we have like bases. Thus, the approach will be the same for each example: repeat the base and add the exponents.
\(\begin{aligned} y^{4} \cdot y^{8} &=y^{4+8} \\ &=y^{12} \end{aligned}\)
\(\begin{aligned} 2^{3} \cdot 2^{5} &=2^{3+5} \\ &=2^{8} \\&=256\end{aligned}\)
\(\begin{aligned}(x+y)^{2}(x+y)^{7} &=(x+y)^{2+7} \\ &=(x+y)^{9} \end{aligned}\)
With a little practice, each of the examples can be simplified mentally. Repeat the base and add the exponents in your head: \(y^{4} \cdot y^{8}=y^{12}, 2^{3} \cdot 2^{5}=2^{8}\) and \((x+y)^{2}(x+y)^{7}=(x+y)^{9}\).
\(3^{4} \cdot 3^{2}\)
- Answer
-
\(729\)
Simplify: \(\left(a^{6} b^{4}\right)\left(a^{3} b^{2}\right)\)
Solution
We’ll use the commutative and associative properties to change the order of operation, then repeat the common bases and add the exponents.
\[\begin{aligned} \left(a^{6} b^{4}\right)\left(a^{3} b^{2}\right) &= a^{6} b^{4} a^{3} b^{2} \quad \color {Red} \text { The associative property allows us to regroup in the order we prefer. } \\ &= a^{6} a^{3} b^{4} b^{2} \quad \color {Red} \text { The commutative property allows us to change the order of multiplication. } \\ &= a^{9} b^{6} \quad \color {Red} \text { Repeat the common bases and add the exponents. } \end{aligned} \nonumber \]
With practice, we realize that if all of the operators are multiplication, then we can multiply in the order we prefer, repeating the common bases and adding the exponents mentally: \(\left(a^{6} b^{4}\right)\left(a^{3} b^{2}\right)=a^{9} b^{6}\).
\(\left(x^{2} y^{6}\right)\left(x^{4} y^{3}\right)\)
- Answer
-
\(x^{6} y^{9}\)
Simplify: \(x^{n+3} \cdot x^{3-2 n}\)
Solution
Again, we repeat the base and add the exponents.
\[\begin{aligned} x^{n+3} \cdot x^{3-2 n} &= x^{(n+3)+(3-2 n)} \quad \color {Red}\text { Repeat the base, add the exponents. } \\ &= x^{6-n} \quad \color {Red} \text { Simplify. Combine like terms. } \end{aligned} \nonumber \]
\(x^{5-n} \cdot x^{4 n+2}\)
- Answer
-
\(x^{3 n+7}\)
Dividing With Like Bases
Like multiplication, we will also be frequently asked to divide exponential expressions with like bases, such as \(x^7/x^4\). Again, the key is to remember that the exponent tells us how many times to write the base as a factor, so we can write:
\[\begin{aligned} \dfrac{x^{7}}{x^{4}} &= \dfrac{x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x}{x \cdot x \cdot x \cdot x} \\ &= \dfrac{\not {x} \cdot \not {x} \cdot \not{x} \cdot \not {x} \cdot x \cdot x \cdot x}{\not {x} \cdot \not {x} \cdot \not {x} \cdot \not {x}} \\ &= x^3 \end{aligned} \nonumber\]
Note how we cancel four \(x’s\) in the numerator for four \(x’s\) in the denominator. However, in a sense we are “subtracting four \(x’s\)” from the numerator, so a faster way to proceed is to repeat the base and subtract the exponents, as follows:
\[\begin{aligned} \dfrac{x^{7}}{x^{4}} &=x^{7-4} \\ &=x^{3} \end{aligned} \nonumber \]
The preceding discussion is an example of the second general property of exponents.
To divide two exponential expressions with like bases, repeat the base and subtract the exponents. Given \(a \neq 0\),
\[\dfrac{a^{m}}{a^{n}}=a^{m-n} \nonumber \]
Note that the subtraction of the exponents follows the rule “top minus bottom.”
Simplify each of the following expressions:
- \(\frac{x^{12}}{x^{3}}\)
- \(\frac{5^{7}}{5^{7}}\)
- \(\frac{(2 x+1)^{8}}{(2 x+1)^{3}}\)
Solution
In each example we have like bases. Thus, the approach will be the same for each example: repeat the base and subtract the exponents.
\(\begin{aligned} \dfrac{x^{12}}{x^{3}} &=x^{12-3} \\ &=x^{9} \end{aligned}\)
\(\begin{aligned} \dfrac{5^{7}}{5^{7}} &=5^{7-7} \\ &=5^{0} \\ &=1 \end{aligned}\)
\(\begin{aligned} \dfrac{(2 x+1)^{8}}{(2 x+1)^{3}} &=(2 x+1)^{8-3} \\ &=(2 x+1)^{5} \end{aligned}\)
With a little practice, each of the examples can be simplified mentally. Repeat the base and subtract the exponents in your head: \(x^{12} / x^{3}=x^{9}, 5^{7} / 5^{4}=5^{3}\) and \((2 x+1)^{8} /(2 x+1)^{3}=(2 x+1)^{5}\).
\(\dfrac{4^{5}}{4^{3}}\)
- Answer
-
\(16\)
Simplify: \(\dfrac{12 x^{5} y^{7}}{4 x^{3} y^{2}}\)
Solution
We first express the fraction as a product of three fractions, the latter two with a common base. In the first line of the following solution, note that if you multiply numerators and denominators of the three separate fractions, the product equals the original fraction on the left.
\[\begin{aligned} \dfrac{12 x^{5} y^{7}}{4 x^{3} y^{2}} &= \dfrac{12}{4} \cdot \dfrac{x^{5}}{x^{3}} \cdot \dfrac{y^{7}}{y^{2}} \quad \color {Red} \text { Break into a product of three fractions. } \\ &= 3 x^{5-3} y^{7-2} \quad \color {Red} \text { Simplify: } 12 / 4=3 . \text { Then repeat the common } \\ &= 3 x^{2} y^{5} \quad \color {Red} \text { Simplify. } \end{aligned} \nonumber \]
Simplify: \(\dfrac{15 a^{6} b^{9}}{3 a b^{5}}\)
- Answer
-
\(5 a^{5} b^{4}\)
Simplify: \(\dfrac{x^{5 n-4}}{x^{3-2 n}}\)
Solution
Again, we repeat the base and subtract the exponents.
\[\begin{aligned} \dfrac{x^{5 n-4}}{x^{3-2 n}} &= x^{(5 n-4)-(3-2 n)} \quad \color {Red} \text { Repeat the base, subtract exponents. } \\ &= x^{5 n-4-3+2 n} \quad \color {Red} \text { Distribute the minus sign. } \\ &= x^{7 n-7} \quad \color {Red} \text { Simplify. Combine like terms. } \end{aligned} \nonumber \]
Simplify: \(\dfrac{x^{3 n-6}}{x^{n+2}}\)
- Answer
-
\(x^{2 n-8}\)
Raising a Power to a Power
Suppose we have an exponential expression raised to a second power, such as \((x^2)^3\). The second exponent tells us to write \(x^2\) as a factor three times:
\[\begin{aligned} \left(x^{2}\right)^{3} &= x^{2} \cdot x^{2} \cdot x^{2} \quad \color {Red} \text { Write } x^{2} \text { as a factor three times. } \\ &= x^{6} \quad \color {Red} \text { Repeat the base, add the exponents. } \end{aligned} \nonumber \]
Note how we added \(2 + 2 + 2\) to get \(6\). However, a much faster way to add “three twos” is to multiply: \(3 \cdot 2=6\). Thus, when raising a “power to a second power,” repeat the base and multiply the exponents, as follows:
\[\begin{aligned}\left(x^{2}\right)^{3} &=x^{2 \cdot 3} \\ &=x^{6} \end{aligned} \nonumber \]
The preceding discussion gives rise to the following third property of exponents.
When raising a power to a power, repeat the base and multiply the exponents. In symbols:
\[(a^m)^n = a^{mn} \nonumber \]
Note that juxtaposing two variables, as in \(mn\), means “\(m\) times \(n\).”
Simplify each of the following expressions:
- \(\left(z^{3}\right)^{5}\)
- \(\left(7^{3}\right)^{0}\)
- \(\left[(x-y)^{3}\right]^{6}\)
Solution
In each example we are raising a power to a power. Hence, in each case, we repeat the base and multiply the exponents.
\(\begin{aligned}\left(z^{3}\right)^{5} &=z^{3.5} \\ &=z^{15} \end{aligned}\)
\(\begin{aligned}\left(7^{3}\right)^{0} &=7^{3.0} \\ &=7^{0} \\ &=1 \end{aligned}\)
\(\begin{aligned}\left[(x-y)^{3}\right]^{6} &=(x-y)^{3 \cdot 6} \\ &=(x-y)^{18} \end{aligned}\)
With a little practice, each of the examples can be simplified mentally. Repeat the base and multiply the exponents in your head: \(\left(z^{3}\right)^{5}=z^{15},\left(7^{3}\right)^{4}=7^{12}\) and \(\left[(x-y)^{3}\right]^{6}=(x-y)^{18}\).
Simplify: \(\left(2^{3}\right)^{4}\)
- Answer
-
\(4096\)
Simplify: \(\left(x^{2 n-3}\right)^{4}\)
Solution
Again, we repeat the base and multiply the exponents.
\[\begin{aligned} \left(x^{2 n-3}\right)^{4} &= x^{4(2 n-3)} \quad \color {Red} \text { Repeat the base, multiply exponents. } \\ &= x^{8 n-12} \quad \color {Red} \text { Distribute the } 4 . \end{aligned} \nonumber \]
Simplify: \(\left(a^{2-n}\right)^{3}\)
- Answer
-
\(a^{6-3 n}\)
Raising a Product to a Power
We frequently have need to raise a product to a power, such as \((xy)^3\). Again, remember the exponent is telling us to write \(xy\) as a factor three times, so:
\[\begin{aligned} (x y)^{3} &=(x y)(x y)(x y) \quad \color {Red} \text {Write } xy \text { as a factor three times.}\\ &=x y x y x y \quad \color {Red} \text {The associative property allows us to group as we please.}\\ &=x x x y y y \quad \color {Red} \text {The commutative property allows us to change the order as we please.}\\ &=x^{3} y^{3} \quad \color {Red} \text {Invoke the exponent definition: } x x x=x^{3} \text { and } y y y=y^{3} \end{aligned} \nonumber \]
However, it is much simpler to note that when you raise a product to a power, you raise each factor to that power. In symbols: \[(xy)^3 = x^3y^3 \nonumber \]
The preceding discussion leads us to a fourth property of exponents.
To raise a product to a power, raise each factor to that power. In symbols:
\[(ab)^n = a^nb^n \nonumber \]
Simplify each of the following expressions:
- \((y z)^{5}\)
- \((-3 y)^{2}\)
Solution
In each example we are raising a product to a power. Hence, in each case, we raise each factor to that power.
\((y z)^{5}=y^{5} z^{5}\)
\(\begin{aligned}(-3 y)^{2} &=(-3)^{2} y^{2} \\ &=9 y^{2} \end{aligned}\)
With a little practice, each of the examples can be simplified mentally. Raise each factor to the indicated power in your head: \((y z)^{5}=y^{5} z^{5} and \((-3 y)^{2}=9 y^{2}\)
Simplify: \((-2 b)^{4}\)
- Answer
-
\(16 b^{4}\)
When raising a product of three factors to a power, it is easy to show that we should raise each factor to the indicated power. For example, \((a b c)^{3}=a^{3} b^{3} c^{3}\). In general, this is true regardless of the number of factors. When raising a product to a power, raise each of the factors to the indicated power.
Simplify: \(\left(-2 a^{3} b^{2}\right)^{3}\)
Solution
Raise each factor to the third power, then simplify.
\[\begin{aligned} \left(-2 a^{3} b^{2}\right)^{3} &= (-2)^{3}\left(a^{3}\right)^{3}\left(b^{2}\right)^{3} \quad \color {Red} \text { Raise each factor to the third power. } \\ &= -8 a^{9} b^{6} \quad \color {Red} \text {Simplify: } (-2)^{3}=-8 \text {. In the remaining factors, raising a power to a power requires that we multiply the exponents. } \end{aligned} \nonumber \]
Simplify: \(\left(-3 x y^{4}\right)^{5}\)
- Answer
-
\(-243 x^{5} y^{20}\)
Simplify: \(\left(-2 x^{2} y\right)^{2}\left(-3 x^{3} y\right)\)
Solution
In the first grouped product, raise each factor to the second power.
\[\begin{aligned} \left(-2 x^{2} y\right)^{2}\left(-3 x^{3} y\right) &=\left((-2)^{2}\left(x^{2}\right)^{2} y^{2}\right)\left(-3 x^{3} y\right) \quad \color {Red} \text { Raise each factor in the first grouped product to the second power.} \\ &=\left(4 x^{4} y^{2}\right)\left(-3 x^{3} y\right) \quad \color {Red} \text { Simplify: } (-2)^{2}=4 \text { and } \left(x^{2}\right)^{2}=x^{4} \end{aligned} \nonumber \]
The associative and commutative property allows us to multiply all six factors in the order that we please. Hence, we’ll multiply \(4\) and \(−3\), then \(x^4\) and \(x^3\), and \(y^2 and y, in that order. In this case, we repeat the base and add the exponents.
\[\begin{aligned} &=-12x^7y^3 \quad \color {Red} \text { Simplify: } (4)(-3)=-12. \text { Also, } x^{4}x^3=x^{7} \text { and } y^2y = y^3\end{aligned} \nonumber\]
Simplify: \(\left(a^{3} b^{2}\right)^{3}\left(-2 a^{2} b^{4}\right)^{2}\)
- Answer
-
\(4 a^{13} b^{14}\)
Raising a Quotient to a Power
Raising a quotient to a power is similar to raising a product to a power. For example, raising \((x/y)^3\) requires that we write \(x/y\) as a factor three times.
\[\begin{aligned}\left(\frac{x}{y}\right)^{3} &=\frac{x}{y} \cdot \frac{x}{y} \cdot \frac{x}{y} \\ &=\frac{x \cdot x \cdot x}{y \cdot y \cdot y} \\ &=\frac{x^{3}}{y^{3}} \end{aligned} \nonumber \]
However, it is much simpler to realize that when you raise a quotient to a power, you raise both numerator and denominator to that power. In symbols:
\[\left(\dfrac{x}{y}\right)^{3}=\frac{x^{3}}{y^{3}} \nonumber \]
This leads to the fifth and final property of exponents.
To raise a quotient to a power, raise both numerator and denominator to that power. Given \(b \neq 0\),
\[\left(\dfrac{a}{b}\right)^{n}=\dfrac{a^{n}}{b^{n}} \nonumber \]
Simplify each of the following expressions:
- \(\left(\dfrac{2}{3}\right)^{2}\)
- \(\left(\dfrac{x}{3}\right)^{3}\)
- \(\left(-\dfrac{2}{y}\right)^{4}\)
Solution
In each example we are raising a quotient to a power. Hence, in each case, we raise both numerator and denominator to that power.
\(\begin{aligned}\left(\dfrac{2}{3}\right)^{2} &=\dfrac{2^{2}}{3^{2}} \\ &=\dfrac{4}{9} \end{aligned}\)
\(\begin{aligned}\left(\dfrac{x}{3}\right)^{3} &=\dfrac{x^{3}}{3^{3}} \\ &=\dfrac{x^{3}}{27} \end{aligned}\)
\(\begin{aligned}\left(-\dfrac{2}{y}\right)^{4} &=\dfrac{2^{4}}{y^{4}} \\ &=\dfrac{16}{y^{4}} \end{aligned}\)
Note that in example (3), raising a negative base to an even power produces a positive result. With a little practice, each of the examples can be simplified mentally. Raise numerator and denominator to the indicated power in your head: \((2 / 3)^{2}=4 / 9,(x / 3)^{3}=x^{3} / 27,\) and \((-2 / y)^{4}=16 / y^{4}\)
Simplify: \(\left(\dfrac{5}{4}\right)^{3}\)
- Answer
-
\(\dfrac{125}{64}\)
Simplify: \(\left(\dfrac{2 x^{5}}{y^{3}}\right)^{2}\)
Solution
Raise both numerator and denominator to the second power, then simplify:
\[\begin{aligned} \left(\dfrac{2 x^{5}}{y^{3}}\right)^{2}=\dfrac{\left(2 x^{5}\right)^{2}}{\left(y^{3}\right)^{2}} \quad \color {Red} \text {Raise numerator and denominator to the second power.} \end{aligned} \nonumber\]
In the numerator, we need to raise each factor of the product to the second power. Then we need to remind ourselves that when we raise a power to a power, we multiply the exponents.
\[\begin{aligned} &= \dfrac{2^{2}\left(x^{5}\right)^{2}}{\left(y^{3}\right)^{2}} \quad \color {Red} \text {Raise each factor in the numerator and denominator to the second power.} \\ &= \dfrac{4 x^{10}}{y^{6}} \quad \color {Red} \text { Simplify: } 2^{2}=4,\left(x^{5}\right)^{2}=x^{10} \text {, and } (y^3)^2=y^6 \end{aligned} \nonumber\]
Simplify: \(\left(\dfrac{a^{4}}{3 b^{2}}\right)^{3}\)
- Answer
-
\(\dfrac{a^{12}}{27 b^{6}}\)


6.4: Infinite geometric series
Elementary mathematics is predominantly about equations and identities. But it is often impossible to capture important mathematical relations in the form of exact equations. This is one reason why inequalities become more central as we progress; another reason is because inequalities allow us to make precise statements about certain infinite processes.
One of the simplest infinite process arises in the formula for the “sum” of an infinite geometric series:
(for ever).
Despite the use of the familiar-looking “+” signs, this can be no ordinary addition. Ordinary addition is defined for two summands; and by repeating the process, we can add three summands (thanks in part to the associative law of addition). We can then add four, or any finite number of summands. But this does not allow us to “add” infinitely many terms as in the above sum. To get round this we combine ordinary addition (of finitely many terms) and simple inequalities to find a new way of giving a meaning to the above “endless sum”. In Problem 116 you used the factorisation
to derive the closed formula:
.
This formula for the sum of a finite geometric series can be rewritten in the form
At first sight, this may not look like a clever move! However, it separates the part that is independent of n from the part on the RHS that depends on n ; and it allows us to see how the second part behaves as n gets large:
when , successive powers of r get smaller and smaller and converge rapidly towards 0,
so the above form of the identity may be interpreted as having the form:
— (an “error term”).
Moreover if
, then the “error term” converges towards 0 as
. In particular, if
, the error term is always positive, so we have proved, for all
and
the difference between the two sides tends rapidly to 0 as .
We then make the natural (but bold) step to interpret this, when , as offering a new definition which explains precisely what is meant by the endless sum
(for ever) ,
declaring that, when ,
( for ever ) = .
More generally, if we multiply every term by a , we see that
ar 2 + ar 3 + ⋯ (for ever) = .
Problem 243 Interpret the recurring decimal 0.037037037 ··· (for ever) as an infinite geometric series, and hence find its value as a fraction.
Problem 244 Interpret the following endless processes as infinite geometric series.
(a) A square cake is cut into four quarters, with two perpendicular cuts through the centre, parallel to the sides. Three people receive one quarter each - leaving a smaller square piece of cake. This smaller piece is then cut in the same way into four quarters, and each person receives one (even smaller) piece - leaving an even smaller residual square piece, which is then cut in the same way. And so on for ever. What fraction of the original cake does each person receive as a result of this endless process?
(b) I give you a whole cake. Half a minute later, you give me half the cake back. One quarter of a minute later, I return one quarter of the cake to you. One eighth of a minute later you return one eighth of the cake to me. And so on. Adding the successive time intervals, we see that
(for ever) = 1,
so the whole process is completed in exactly 1 minute. How much of the cake do I have at the end, and how much do you have?
Problem 245 When John von Neumann (1903-1957) was seriously ill in hospital, a visitor tried (rather insensitively) to distract him with the following elementary mathematics problem.
Have you heard the one about the two trains and the fly? Two trains are on a collision course on the same track, each travelling at 30 km/h. A super-fly starts on Train A when the trains are 120 km apart, and flies at a constant speed of 50 km/h - from Train A to Train B, then back to Train A, and so on. Eventually the two trains collide and the fly is squashed. How far did the fly travel before this sad outcome?


10.3: The Importance of Studying Communication in Groups and Teams
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One of the reasons communication scholars study groups and teams is because of the overwhelming amount of time we spend interacting in groups in professional contexts. More and more professional organizations are turning to groups and teams as an essential way of conducting business and getting things done. Even professions that are seemingly independent, such as being a college professor, are heavily laden with group work. The process of writing this book was a group effort as the authors and their students worked in groups to bring the book to you. Each of us had specific roles and tasks to perform.
Another vital area of group communication concerns the study of social change or social movement organizations. Groups such as People for the Ethical Treatment of Animals (PETA), the National Association for the Advancement of Colored People (NAACP), and the National Organization of Women (NOW) are all groups bound together by a shared social and political commitment—-to promote the rights of nonhuman animals, African-Americans, and women respectively. While individuals can be committed to these ideas, the social, political, and legal rights afforded to groups like these would not have been possible through individual action alone. It was when groups of like-minded people came together with shared commitments and goals, pooling their skills and resources, that change occurred. Just about any Interest Group you can think of has a presence in Washington D.C. and spends money to maintain that presence. Visit here for more information on Interest Groups .
The study of social movements reveals the importance of groups for accomplishing goals. Bowers, Ochs, Jensen and Schulz, in The Rhetoric of Agitation and Control , explain seven progressive and cumulative strategies through which movements progress as they move toward success. Three of the seven strategies focus explicitly on group communication—-promulgation, solidification, and polarization. Promulgation refers to the “a strategy where agitators publicly proclaim their goals and it includes tactics designed to win public support” (23). Without a sufficient group the actions of individual protestors are likely to be dismissed. The strategy of solidification “occurs mainly inside the agitating group” and is “primarily used to unite followers” (29). The point is to unite group members and provide sufficient motivation and support. The communication that occurs through the collective action of singing songs or chanting slogans serves to unite group members. Because the success of social movements depends in part on the ability to attract a large number of followers, most employ the strategy of polarization , which is designed to persuade neutral individuals or “fence sitters” to join a group (40). The essence of this strategy is captured in the quote from Eldridge Cleaver, “You are either part of the problem or part of the solution.” Taken together these three strategies stress that the key to group success is the sustained effort of group members working together through communication.
Not only do Communication scholars focus on work and social movements, we are also interested in the role that one’s cultural identity and membership plays in our communicative choices, and how we interpret the communication of others. This focus sheds interesting insights when we examine membership and communication in groups and teams. One reason for this is that different cultures emphasize the role of individuals while other cultures emphasize the importance of the group. For example, collectivist cultures are ones that place high value on group work because they understand that outcomes of our communication impact all members of the community and the community as a whole, not just the individuals in the group . Conversely, individualistic cultures are ones that place high value on the individual person above the needs of the group . Thus, whether we view group work as favorable or unfavorable may stem from our cultural background. The U.S. is considered an individualistic culture in that we value the work and accomplishments of the individual through ideals such as being able to “pull yourself up by the bootstraps” and create success for yourself.
Case In Point
The Historic Number of Women in the 113th Congress
This Congress has 102 female members, including 20 female Senators, the most ever in the history of America. This was also an historic election for women of color. More than half of Asian Americans elected to the 113th Congress are women– including Mazie Hirono (D-HI), who is only the second woman of color and the only Asian American woman ever elected to the Senate. In the House, Grace Meng (D) will become New York’s first Asian American member of Congress, and Iraq War veteran Tulsi Gabbard (D-HI) the first American Samoan elected as a voting member of Congress. They will both join Tammy Duckworth (D-IL), a Thai-born, double amputee war hero. Among Latinas, Congresswoman-elect Michele Lujan Grisham became the first Hispanic woman elected representative from New Mexico.
The Women’s Caucus
The Congressional Caucus for Women’s Issues (CCWI) is a Congressional Member Organization (CMO) registered with the House Organization (CMO) registered with the House Administration Committee; its membership includes the women members of the House of Representatives. CCWI is not formally affiliated with Women’s Policy, Inc.; however, the two organizations share similar goals. This relates to our collectivistic culture in organizational groups.
WPI is a nonprofit, nonpartisan organization whose sole focus is to help ensure that the most informed decisions on key women’s issues are made by policymakers at the federal, state, and local levels. Audiences include elected officials, regulators, women’s groups, labor groups, academia, the business community, the media, and the general public.
WPI achieves and shares its rare quality of insight into relevant issues by researching and producing the best available information in the form of compelling and unbiased legislative analyses, issue summaries, impact assessments, and educational briefings. This ensures that policy decisions affecting women and their families have the benefit of input from the most objective sources possible.
For more info on The Women’s Caucus visit http://www.womenspolicy.org/
Power in Groups
Given the complexity of group interaction, it’s short-sighted to try to understand group communication without looking at notions of power (think back to Critical Theories and Research Methods). Power influences how we interpret the messages of others and determines the extent to which we feel we have the right to speak up and voice our concerns and opinions to others. Take a moment to reflect on the different ways you think about power. What images come to mind for you when you think of power? Are there different kinds of power? Are some people inherently more powerful than others? Do you consider yourself to be a powerful person? We highlight three ways to understand power as it relates to group and team communication. The word “power” literally means “to be able” and has many implications.
If you associate power with control or dominance, this refers to the notion of power as power-over . According to Starhawk, “power-over enables one individual or group to make the decisions that affect others, and to enforce control ” (9). Control can and does take many forms in society. Starhawk explains that,
- This power is wielded from the workplace, in the schools, in the courts, in the doctor’s office. It may rule with weapons that are physical or by controlling the resources we need to live: money, food, medical care; or by controlling more subtle resources: information, approval, love. We are so accustomed to power-over, so steeped in its language and its implicit threats, that we often become aware of its functioning only when we see its extreme manifestations. (9)
When we are in group situations and someone dominates the conversation, makes all of the decisions, or controls the resources of the group such as money or equipment, this is power-over.
Power-from-within refers to a more personal sense of strength or agency. Power-from-within manifests itself when we can stand, walk, and speak “words that convey our needs and thoughts ” (Starhawk 10). In groups, this type of power “arises from our sense of connection, our bonding with other human beings, and with the environment” (10). As Heider explains in The Tao of Leadership , “Since all creation is a whole, separateness is an illusion. Like it or not, we are team players. Power comes through cooperation, independence through service, and a greater self through selflessness” (77). If you think about your role in groups, how have you influenced other group members? Your strategies indicate your sense of power-from-within.
Finally, groups manifest power-with , which is “ the power of a strong individual in a group of equals, the power not to command, but to suggest and be listened to, to begin something and see it happen ” (Starhawk 10). For this to be effective in a group or team, at least two qualities must be present among members: 1) All group members must communicate respect and equality for one another, and 2) The leader must not abuse power-with and attempt to turn it into power-over. Have you ever been involved in a group where people did not treat each others as equals or with respect? How did you feel about the group? What was the outcome? Could you have done anything to change that dynamic?
Obviously, communication is the central activity of every group because it is how we organize and maintain groups. While we can all tell positive and negative stories about being in groups, how are they formed in the first place?
Contributions and Affiliations
- Survey of Communication Study. Authored by : Scott T Paynton and Linda K Hahn. Provided by : Humboldt State University. Located at : https://en.wikibooks.org/wiki/Survey_of_Communication_Study . License : CC BY-SA: Attribution-ShareAlike
- Image of team-building exercise. Authored by : Spaynton. Located at : https://commons.wikimedia.org/wiki/File:Powergroups.png . License : CC BY-SA: Attribution-ShareAlike