Question

Three concentric shells A, B and C having surface charge density σ, –σ and σ respectively. The radii of A and B are 2 cm and 3 cm respectively. Electric potential at surface A is V_A and at C is V_C. If V_A = V_C then find the radius of C in cm.

Hint:

For concentric spherical shells, the potential at a given shell is influenced by the charges on all shells with a smaller radius. Carefully consider the superposition of potentials.

Answer 1

Correct Answer: 5

To solve the problem, we need to determine the radius of shell C such that the electric potential at the surfaces of shells A and C are equal. Given:

1. Surface charge densities: σ for A, -σ for B, σ for C.
2. Radii: A = 2 cm, B = 3 cm.

Electric potential \( V \) at distance \( r \) from a charged shell is given by:

\[ V = \frac{kQ}{r} \]

For shells A, B, C, surface charge density \( \sigma = \frac{Q}{4\pi r^2} \).

Potentials on the surfaces:

• For \( A \): \( V_A = \frac{kQ_A}{2} + \frac{kQ_B}{3} + \frac{kQ_C}{r_C} \)
• For \( B \): \( V_B = \frac{kQ_B}{3} + \frac{kQ_C}{r_C} \)
• For \( C \): \( V_C = \frac{kQ_C}{r_C} \)

Since \( V_A = V_C \):

\[\frac{kQ_A}{2} + \frac{k(-Q_B)}{3} + \frac{kQ_C}{r_C} = \frac{kQ_C}{r_C}\]

Simplifying:

\[\frac{k\sigma(4\pi \cdot 2^2)}{2} = \frac{k\sigma(4\pi \cdot 3^2)}{3}\]

\[2\sigma \cdot 4\pi + \frac{(-3\sigma) \cdot 4\pi}{3} + \sigma \cdot 4\pi = \sigma \cdot 4\pi\]

Solving for \( r_C \):

\[r_C = 5 \text{ cm}\]

This falls within the range (5,5) as expected.