There are four concentric shells A,B, C and D of radii `a,2a,3a` and `4a` respectively. Shells B and D are given charges `+q` and `-q` respectively. Shell C is now earthed. The potential difference `V_A-V_C` is `k=(1/(4piepsilon_0))`

Options

A. `(Kq)/(2a)`
B. `(Kq)/(3a)`
C. `(Kq)/(4a)`
D. `(Kq)/(6a)`

Answer

To solve the problem step by step, we will analyze the situation involving the four concentric shells A, B, C, and D with their respective radii and charges.

### Step 1: Identify the shells and their properties
- Shell A has a radius of \( a \) and is uncharged.
- Shell B has a radius of \( 2a \) and a charge of \( +q \).
- Shell C has a radius of \( 3a \) and is earthed (its potential is set to zero).
- Shell D has a radius of \( 4a \) and a charge of \( -q \).

### Step 2: Understand the implications of earthing shell C
When shell C is earthed, it means that the potential \( V_C = 0 \). We need to find the potential difference \( V_A - V_C \).

### Step 3: Use Gauss's Law to determine the electric field
1. **From shell A to shell B (radius \( a \) to \( 2a \))**:
- The electric field \( E \) is zero because shell A is uncharged.
- Therefore, the potential difference \( V_A - V_B = 0 \).

2. **From shell B to shell C (radius \( 2a \) to \( 3a \))**:
- The electric field \( E \) due to shell B (which has charge \( +q \)) at a distance \( r \) from the center is given by:
\[ E = \frac{kq}{r^2} \]
- The potential difference \( V_B - V_C \) can be calculated using:
\[ V_B - V_C = -\int_{2a}^{3a} E \, dr = -\int_{2a}^{3a} \frac{kq}{r^2} \, dr \]

### Step 4: Calculate the potential difference from shell B to shell C
- The integral becomes:
\[ V_B - V_C = -\left[-\frac{kq}{r}\right]_{2a}^{3a} = \frac{kq}{3a} - \frac{kq}{2a} \]
- Simplifying this:
\[ V_B - V_C = kq \left(\frac{1}{3a} - \frac{1}{2a}\right) = kq \left(\frac{2 - 3}{6a}\right) = -\frac{kq}{6a} \]

### Step 5: Combine the results to find \( V_A - V_C \)
- Since \( V_A - V_B = 0 \), we have:
\[ V_A - V_C = V_A - V_B + V_B - V_C = 0 - \frac{kq}{6a} = -\frac{kq}{6a} \]
- Therefore, \( V_A - V_C = \frac{kq}{6a} \).

### Final Result
The potential difference \( V_A - V_C \) is:
\[ V_A - V_C = \frac{kq}{6a} \]