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, we need to find the potential difference \( V_A - V_C \) given the configuration of the concentric shells and their charges.

### Step-by-Step Solution:

1. **Identify the Charges and Radii**:
- Shell A has radius \( a \) and no charge.
- Shell B has radius \( 2a \) and charge \( +q \).
- Shell C has radius \( 3a \) and is earthed (potential \( V_C = 0 \)).
- Shell D has radius \( 4a \) and charge \( -q \).

2. **Calculate the Potential at Shell C (\( V_C \))**:
Since shell C is earthed, its potential \( V_C = 0 \). The potential at shell C due to the charges on shells B and D can be calculated as follows:
\[ V_C = \frac{kq}{2a} + \frac{-kq}{4a} \]
Here, \( k = \frac{1}{4\pi \epsilon_0} \).

Substituting the values:
\[ V_C = \frac{kq}{2a} - \frac{kq}{4a} = \frac{2kq}{4a} - \frac{kq}{4a} = \frac{kq}{4a} \]

Since \( V_C = 0 \) (because it is earthed), we have:
\[ \frac{kq}{4a} + \frac{kQ'}{3a} = 0 \]
where \( Q' \) is the induced charge on shell C.

3. **Find the Induced Charge on Shell C (\( Q' \))**:
Rearranging the equation gives:
\[ \frac{kQ'}{3a} = -\frac{kq}{4a} \]
Simplifying this, we find:
\[ Q' = -\frac{q}{4} \]

4. **Calculate the Potential at Shell A (\( V_A \))**:
The potential at shell A due to the charges on shells B, C, and D is given by:
\[ V_A = \frac{kq}{2a} + \frac{kQ'}{3a} - \frac{kq}{4a} \]
Substituting \( Q' = -\frac{q}{4} \):
\[ V_A = \frac{kq}{2a} - \frac{kq/4}{3a} - \frac{kq}{4a} \]
Now, substituting the values:
\[ V_A = \frac{kq}{2a} - \frac{kq}{12a} - \frac{kq}{4a} \]
To combine these terms, we need a common denominator (12a):
\[ V_A = \frac{6kq}{12a} - \frac{kq}{12a} - \frac{3kq}{12a} = \frac{(6 - 1 - 3)kq}{12a} = \frac{2kq}{12a} = \frac{kq}{6a} \]

5. **Calculate the Potential Difference (\( V_A - V_C \))**:
Since \( V_C = 0 \):
\[ V_A - V_C = V_A = \frac{kq}{6a} \]

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