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Electrostatic energy of the system

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Figure shows three concentric conducting spherical shells with inner and outer shells earthed and the middle shell is given a charge q. Find the electrostatic energy of the system stored in the region I and II.

I tried as

As the potential of sperical shell I and III are zero , I got two equation
Potential at shell I is zero hence $$K/r[q_1 + (2q/5) + (2q_3 /7)] =0$$
Potential at shell III is zero hence $$2K/7r[q_1 + q + q_3]=0$$

From this I got $q_1 =(-74q/25)$ and $q_3 = (-49q/25)$

asked Feb 6, 2017 in Physics Problems by koolman (4,286 points)
edited Feb 8, 2017 by koolman
I see only one equation here. ... So what do you do next?

Isn't this a trick question? The electric charge on II resides on the outer surface. So no charge is induced on I. There is no electric field between conductors I & II, so no energy is stored in this region.

Do you have a worked solution?
The potential of conductor I should be zero . I assumed charge of conductor I as q$_1$  so there will be charge induced that charge on conductor II would be -q$_1$ on inner side and q+q$_1$ on outer surface.

1 Answer

1 vote
Best answer

The arrangement can be viewed as two spherical capacitors in series.

Assume the charges on the inner and outer surfaces of the middle spherical shell are $q_1$ and $q_2$. The total charge on the middle shell is $q_1+q_2=q$. The charges induced on the inner and outer shells are $-q_1$ and $-q_2$ respectively.

The PD between the inner and outer shell of a spherical capacitor is $kQ(\frac{1}{a}-\frac{1}{b})$ where $Q$ is the charge on the inner shell and $a,b$ are the radii of inner and outer shells. So in this case the PDs between inner and middle shells and between middle and outer shells are
$V_1=k(-q_1)(\frac{1}{r}-\frac{1}{2.5r})= -\frac35 \frac{kq_1}{r}$
$V_2=k(+q_2)(\frac{1}{2.5r}-\frac{1}{3.5r})=\frac{4}{35} \frac{kq_2}{r}$.
These PDs are equal (and opposite in sign) because the middle shell is common and the outer shells are at the same potential, so
$\frac35 \frac{kq_1}{r}=\frac{4}{35} \frac{kq_2}{r}$
$q_2= \frac{21}{4}q_1$.
Substituting into $q_1+q_2=q$ gives

The energies stored in capacitors 1 and 2 are
$U_1=\frac12q_1V_1=\frac{3}{10} \frac{kq_1^2}{r}$
$U_2=\frac12 q_2V_2=\frac{2}{35}\frac{kq_2^2}{r}$

Except for the discrepancy over the sign of $q_1$, which makes no difference to the answers, these expressions agree with the answers you have been given.

answered Feb 8, 2017 by sammy gerbil (28,746 points)
edited Feb 8, 2017 by sammy gerbil
What mistake I have done
Sorry, I don't know, because I do not understand how you got your equations. You have not provided enough detail in your explanation.

Why don't you post your solution (in detail) as an answer?  Then I can comment on it.
Now is it fine
No, I still do not understand how you got these equations.
I have just as equated the total potential at shell I and III to be zero .
How did you get the left hand side of each equation?
I want to know what you did. I do not want to spend time looking at other sites to try to figure out what you did. If you are not prepared to explain in detail what you did, I cannot help.
At shell I potential due to charge q$_1$ is Kq$_1$/r , due to q is 2Kq/5r and due to q$_3$ is 2Kq$_3$/7r  . The same way for shell III.
That is the potential at shell III due to charge q$_1$ , q and q$_3$ are 2Kq$_1$/7r , 2Kq/7r and 2Kq$_3$/7r respectively .

Now I have just added these potential and equated to zero
OK I understand what you have done now. You have used the Principle of Superposition.

What you did wrong is that you did not use the correct sign for $q_2$ at conductor III. You assumed charge $+q_1$ is induced on outer surface of conductor I, charges on conductor II are $-q_1$ on inner surface  and $q+q_1=q_2$ on outer surface, and charge $-q_2$ on inner surface of conductor III. However, in your equations you used $+q_2$ as the charge on conductor III when it should be $-q_2$.

After you correct that mistake you get $q_2=\frac{21}{25}q$ and $q_1=-\frac{4}{25}q$.