The work done in increasing the P.D. across the plates of a capacitor from 4V to 6V is W then work done in increasing the P.D. from 6V to 8V is

Options

A. `(4W)/3`
B. `(3W)/4`
C. `(7W)/5`
D. `(5W)/7`

Answer

To solve the problem, we need to calculate the work done in increasing the potential difference (P.D.) across the plates of a capacitor from 6V to 8V, given that the work done in increasing the P.D. from 4V to 6V is W.

### Step-by-Step Solution:

1. **Understand the Formula for Work Done on a Capacitor**:
The work done (W) in charging a capacitor can be expressed in terms of the change in energy stored in the capacitor. The energy (U) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
where \(C\) is the capacitance and \(V\) is the potential difference.

2. **Calculate Work Done from 4V to 6V**:
The work done in increasing the P.D. from 4V to 6V is:
\[ W = U_2 - U_1 = \left(\frac{1}{2} C (6^2)\right) - \left(\frac{1}{2} C (4^2)\right) \]
Simplifying this, we have:
\[ W = \frac{1}{2} C (36 - 16) = \frac{1}{2} C \cdot 20 = 10C \]

3. **Calculate Work Done from 6V to 8V**:
Now, we need to find the work done in increasing the P.D. from 6V to 8V. This can be expressed as:
\[ W' = U_4 - U_3 = \left(\frac{1}{2} C (8^2)\right) - \left(\frac{1}{2} C (6^2)\right) \]
Simplifying this, we have:
\[ W' = \frac{1}{2} C (64 - 36) = \frac{1}{2} C \cdot 28 = 14C \]

4. **Relate the Two Work Done Values**:
We know from the previous calculation that \(W = 10C\). Therefore, we can express \(W'\) in terms of \(W\):
\[ W' = 14C = \frac{14}{10} W = \frac{7}{5} W \]

### Final Answer:
The work done in increasing the P.D. from 6V to 8V is:
\[ W' = \frac{7}{5} W \]