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)/(4a)`
C. `(kq)/(4a)`
D. `(kq)/(6a)`

Explanation

To solve the problem step by step, we will analyze the potential at different shells due to the charges on the other shells.

### Step 1: Understand the Configuration
We have four concentric shells A, B, C, and D with radii \(a\), \(2a\), \(3a\), and \(4a\) respectively. Shell B has a charge of \(+q\), shell D has a charge of \(-q\), and shell C is earthed (potential = 0).

### Step 2: Calculate the Potential at Shell A
The potential at shell A (\(V_A\)) is due to the charges on shells B, C, and D. The formula for the potential \(V\) due to a shell of charge \(Q\) at a distance \(r\) from its center is given by:
\[ V = k \frac{Q}{r} \]
where \(k = \frac{1}{4\pi \epsilon_0}\).

1. **Potential due to shell B** at radius \(2a\):
\[ V_B = k \frac{q}{2a} \]

2. **Potential due to shell C** (which is earthed, so it has a charge \(Q_C\) that we will find later):
\[ V_C = k \frac{Q_C}{3a} \]

3. **Potential due to shell D** at radius \(4a\):
\[ V_D = -k \frac{q}{4a} \]

Combining these, the total potential at shell A is:
\[ V_A = k \frac{q}{2a} + k \frac{Q_C}{3a} - k \frac{q}{4a} \]

### Step 3: Calculate the Potential at Shell C
Since shell C is earthed, its potential \(V_C\) is:
\[ V_C = 0 \]

### Step 4: Solve for \(Q_C\)
Since \(V_C = 0\), we can express the potential at shell C:
\[ 0 = k \frac{q}{3a} + k \frac{Q_C}{3a} - k \frac{q}{4a} \]
Multiplying through by \(3a\) to eliminate the denominators:
\[ 0 = q + Q_C - \frac{3q}{4} \]
Rearranging gives:
\[ Q_C = -q + \frac{3q}{4} = -\frac{q}{4} \]

### Step 5: Substitute \(Q_C\) Back into \(V_A\)
Now substituting \(Q_C = -\frac{q}{4}\) back into the expression for \(V_A\):
\[ V_A = k \frac{q}{2a} + k \frac{-\frac{q}{4}}{3a} - k \frac{q}{4a} \]
Calculating each term:
1. The first term: \(k \frac{q}{2a}\)
2. The second term: \(k \frac{-\frac{q}{4}}{3a} = -k \frac{q}{12a}\)
3. The third term: \(-k \frac{q}{4a}\)

Combining these:
\[ V_A = k \left( \frac{q}{2a} - \frac{q}{12a} - \frac{3q}{12a} \right) \]
\[ = k \left( \frac{6q}{12a} - \frac{4q}{12a} \right) = k \frac{2q}{12a} = k \frac{q}{6a} \]

### Step 6: Calculate the Potential Difference \(V_A - V_C\)
Since \(V_C = 0\):
\[ V_A - V_C = V_A - 0 = V_A = k \frac{q}{6a} \]

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