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

Step 1: Write the Potential at Point \( y \)

The potential \( V_y \) at point \( y \) is the sum of potentials due to charges \( q_x \), \( q_y \), and \( q_z \):

\[ V_y = \frac{q_x}{4\pi\varepsilon_0 a} + \frac{q_y}{4\pi\varepsilon_0 b} + \frac{q_z}{4\pi\varepsilon_0 c} \]

Substitute the given charges:

\[ q_x = \sigma 4\pi a^2, \, q_y = -\sigma 4\pi b^2, \, q_z = \sigma 4\pi c^2 \]

\[ V_y = \frac{\sigma 4\pi a^2}{4\pi\varepsilon_0 a} - \frac{\sigma 4\pi b^2}{4\pi\varepsilon_0 b} + \frac{\sigma 4\pi c^2}{4\pi\varepsilon_0 c} \]

Simplify each term:

\[ V_y = \frac{\sigma a}{\varepsilon_0} - \frac{\sigma b}{\varepsilon_0} + \frac{\sigma c}{\varepsilon_0} \]

Combine terms:

\[ V_y = \frac{\sigma}{\varepsilon_0} (a - b + c) \]

Step 2: Relate \( c \) to \( a \) and \( b \)

Using the condition \( c(a - b + c) = a^2 - b^2 + c^2 \), expand and simplify:

\[ c(a - b) + c^2 = a^2 - b^2 + c^2 \]

Cancel \( c^2 \):

\[ c(a - b) = (a + b)(a - b) \]

Factorize:

\[ c = a + b \]

Step 3: Solve for \( c \)

Substitute \( a = 2 \, \text{cm} \) and \( b = 3 \, \text{cm} \):

\[ c = a + b = 2 + 3 = 5 \, \text{cm} \]

Final Answer:

The value of \( c \) is:

\( c = 5 \, \text{cm}. \)