The work done against electric in increasing the potential difference of a condenser from 20 V to 40 V is W. The work done in increasing its potential difference from 40 V to 50 V will be

Options

A. 4W
B. `(3W)/(4)`
C. 2W
D. `(W)/(2)`

Answer

To solve the problem, we need to calculate the work done in two different scenarios of increasing the potential difference across a capacitor (or condenser).

### Step-by-Step Solution:

1. **Understanding the Work Done Formula**:
The work done (W) in charging a capacitor when the potential difference is increased from \( V_1 \) to \( V_2 \) is given by the formula:
\[ W = \frac{1}{2} C (V_2^2 - V_1^2) \]
where \( C \) is the capacitance of the capacitor.

2. **Calculating Work Done from 20V to 40V**:
In the first case, we increase the potential difference from \( V_1 = 20V \) to \( V_2 = 40V \):
\[ W = \frac{1}{2} C (40^2 - 20^2) \]
\[ = \frac{1}{2} C (1600 - 400) = \frac{1}{2} C (1200) = 600C \]
We are given that this work done is \( W \). Therefore, we can equate:
\[ W = 600C \quad \text{(Equation 1)} \]

3. **Calculating Work Done from 40V to 50V**:
In the second case, we increase the potential difference from \( V_1 = 40V \) to \( V_2 = 50V \):
\[ W' = \frac{1}{2} C (50^2 - 40^2) \]
\[ = \frac{1}{2} C (2500 - 1600) = \frac{1}{2} C (900) = 450C \]

4. **Relating the Two Work Done Values**:
We can express \( C \) in terms of \( W \) from Equation 1:
\[ C = \frac{W}{600} \]
Now substituting this into the expression for \( W' \):
\[ W' = 450C = 450 \left(\frac{W}{600}\right) = \frac{450W}{600} = \frac{3W}{4} \]

5. **Conclusion**:
Therefore, the work done in increasing the potential difference from 40V to 50V is:
\[ W' = \frac{3W}{4} \]

### Final Answer:
The work done in increasing the potential difference from 40V to 50V will be \( \frac{3W}{4} \).

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