A capacitor of capacitance `5muF` is charged with `5muC` charge its capacitance is changed to `2muF` by some external agent. The work done by external agent is

Options

A. `40.5xx10^(-7)J`
B. `42.5xx10^(-7)J`
C. `37.5xx10^(-7)J`
D. `30.5xx10^(-7)J`

Answer

To find the work done by the external agent when changing the capacitance of a charged capacitor, we can follow these steps:

### Step 1: Understand the Initial Conditions
We have a capacitor with:
- Initial capacitance, \( C_i = 5 \, \mu F = 5 \times 10^{-6} \, F \)
- Charge on the capacitor, \( Q = 5 \, \mu C = 5 \times 10^{-6} \, C \)

### Step 2: Calculate the Initial Energy
The energy stored in a capacitor is given by the formula:
\[ U = \frac{Q^2}{2C} \]
Substituting the initial values:
\[ U_i = \frac{(5 \times 10^{-6})^2}{2 \times (5 \times 10^{-6})} = \frac{25 \times 10^{-12}}{10 \times 10^{-6}} = \frac{25}{10} \times 10^{-6} = 2.5 \times 10^{-6} \, J \]

### Step 3: Understand the Final Conditions
The capacitance is changed to:
- Final capacitance, \( C_f = 2 \, \mu F = 2 \times 10^{-6} \, F \)

### Step 4: Calculate the Final Energy
Using the same energy formula for the final state:
\[ U_f = \frac{Q^2}{2C_f} = \frac{(5 \times 10^{-6})^2}{2 \times (2 \times 10^{-6})} = \frac{25 \times 10^{-12}}{4 \times 10^{-6}} = \frac{25}{4} \times 10^{-6} = 6.25 \times 10^{-6} \, J \]

### Step 5: Calculate the Work Done by the External Agent
The work done by the external agent is the change in energy:
\[ W = U_f - U_i = 6.25 \times 10^{-6} - 2.5 \times 10^{-6} = 3.75 \times 10^{-6} \, J \]

### Step 6: Convert Work Done to the Required Format
To express this in the required format:
\[ W = 3.75 \times 10^{-6} \, J = 37.5 \times 10^{-7} \, J \]

### Final Answer
The work done by the external agent is:
\[ \boxed{37.5 \times 10^{-7} \, J} \]

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