
Albert Einstein

Saturday, January 14, 2012

AP Physics C - Answer to Free Response Practice Question on Electrostaics

"I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend".
–Sir Isaac Newton

In the post dated 11th January 2012, a free-response question for practice was given to you. As promised, I give below a model answer. The question also is given for convenience:






[This question was asked in a different form in IIT entrance examination earlier].

(a) The disc can be imagine to be made of a large number of concentric rings of radii ranging from zero to a. To derive an expression for the potential at a point P (Fig.) on the axis of the disc at distance x from the centre, consider one such ring of radius r and width dr as shown in the figure. The total charge on the ring is rdrσ since the area of the top surface of the ring is rdr.

All charges on the ring are at the same distance √(r2 + x2) from the point P so that the electric potential at P due to the charges on the ring is (1/4πε0){rdrσ/√(r2 + x2)} where ε0 is the permittivity of free space.

The electric potential at P due to the charges on the entire disc [(P)] is obtained by adding the contributions of all the rings forming the entire disc. Therefore we have

V(P) = 0a [(1/4πε0){rdrσ/√(r2 + x2)}]dr

Or, V(P) = (σ/2ε0) 0a [r/√(r2 + x2)] dr

Since the value of the integral is √(r2 + x2) and the upper and lower limits (of r) are a and zero, we get

V(P) = (σ/2ε0)[√(a2 + x2) x]

(b) The electric potential at the centre of the disc [V(O)] is obtained by putting x = 0 in the above general expression for the potential on the axis.

Therefore, V(O) = σa/2ε0

(c) When the particle falls from height d and just comes to rest (momentarily) at O, as is evident from the question, it gains electric potential energy at the cost of gravitational potential energy. Therefore, by the law of conservation of energy we have

mgd =q[V(O) V(P)] where V(P) here is the electric potential energy when the point P is at height d and is equal to (σ/2ε0){√(a2 + d2) d}. Therefore we have

mgd =q[(σa/2ε0) – (σ/2ε0){√(a2 + d2) d}]

Or, mgd =(qσ/2ε0) [a –{√(a2 + d2) d}]

Therefore, d =(q/m)(σ/2ε0g) [a –{√(a2 + d2) d}]

Since the specific charge q/m =8ε0g/σ, as given in the question, we get

d =(8ε0g/σ)(σ/2ε0g) [a –{√(a2 + d2) d}] = 4[a –{√(a2 + d2) d}]

Or, 4√(a2 + d2) = 3d + 4a

Squaring, 16a2 + 16d2 = 9d2 + 16a2 + 24ad

Or, 7d2 = 24ad from which d = 24a/7

(d) Initially the particle moves down with an acceleration because of the gravitational pull mg even though the electrostatic repulsive force due to the like charges on the disc opposes the motion. As the particle approaches the disc, the electric force increases where as the gravitational force remains constant. But the particle overshoots its final equilibrium position because of inertia. The downward motion of the particle gets decelerated because of the increasing electric repulsive force and it comes to rest momentarily at the central point O. The particle then moves upwards because of the electric repulsive force which is greater in magnitude here, compared to the gravitational pull. The particle oscillates about its final equilibrium position and finally comes to rest at the final equilibrium position which is at a height from the point O.

1 comment:

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