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))`

Answer

To solve the problem, we need to calculate the potential difference \( V_A - V_C \) between the shells A and C, given the conditions of the concentric shells and their charges.

### Step-by-Step Solution:

1. **Identify the Shells and Their Charges**:
- Shell A has radius \( a \).
- Shell B has radius \( 2a \) and charge \( +q \).
- Shell C has radius \( 3a \) and is earthed (potential is set to zero).
- Shell D has radius \( 4a \) and charge \( -q \).

2. **Understand the Effect of Earthing Shell C**:
- When shell C is earthed, its potential \( V_C \) becomes zero.

3. **Use Gauss's Law to Determine Electric Fields**:
- For the region between shells A and B (from \( r = a \) to \( r = 2a \)), the electric field \( E \) is zero because the charge enclosed is zero.
- From shell B to shell C (from \( r = 2a \) to \( r = 3a \)), the electric field can be calculated using Gauss's Law:
\[ E = \frac{kq}{r^2} \]
where \( k = \frac{1}{4\pi\epsilon_0} \).

4. **Calculate the Potential Difference \( V_A - V_C \)**:
- The potential difference can be expressed as:
\[ V_A - V_C = -\int_{r_A}^{r_C} E \, dr \]
- Here, \( r_A = a \) and \( r_C = 3a \). The potential difference can be split into two parts:
\[ V_A - V_C = -\int_{a}^{2a} 0 \, dr - \int_{2a}^{3a} \frac{kq}{r^2} \, dr \]
- The first integral evaluates to zero since the electric field is zero between shells A and B.

5. **Evaluate the Second Integral**:
- The second integral becomes:
\[ -\int_{2a}^{3a} \frac{kq}{r^2} \, dr = -kq \left[-\frac{1}{r}\right]_{2a}^{3a} \]
- Evaluating this gives:
\[ -kq \left(-\frac{1}{3a} + \frac{1}{2a}\right) = kq \left(\frac{1}{2a} - \frac{1}{3a}\right) \]
- Finding a common denominator (6a):
\[ = kq \left(\frac{3}{6a} - \frac{2}{6a}\right) = \frac{kq}{6a} \]

6. **Final Result**:
- Therefore, the potential difference \( V_A - V_C \) is:
\[ V_A - V_C = \frac{kq}{6a} \]

### Conclusion:
The potential difference \( V_A - V_C \) is \( \frac{kq}{6a} \).