A variable condenser is permanently connect to a `100V` battery. If the capacity is charged from `2muF` to `10muF`, then change in energy is equal to

Options

A. `2 xx 10^(-2) J`
B. `2.5 xx 10^(-2) J`
C. `3.5 xx 10^(-2)J`
D. `4 xx 10^(-2) J`

Answer

To find the change in energy of a variable condenser connected to a 100V battery when the capacitance changes from 2μF to 10μF, we can follow these steps:

### Step 1: Understand the formula for energy stored in a capacitor
The energy (U) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
where:
- \( U \) is the energy stored,
- \( C \) is the capacitance,
- \( V \) is the voltage across the capacitor.

### Step 2: Calculate the initial energy with \( C_1 = 2μF \)
Given:
- \( C_1 = 2μF = 2 \times 10^{-6} F \)
- \( V = 100V \)

Using the formula:
\[ U_1 = \frac{1}{2} C_1 V^2 \]
\[ U_1 = \frac{1}{2} \times (2 \times 10^{-6}) \times (100)^2 \]
\[ U_1 = \frac{1}{2} \times (2 \times 10^{-6}) \times 10000 \]
\[ U_1 = \frac{1}{2} \times 2 \times 10^{-2} \]
\[ U_1 = 1 \times 10^{-2} J \]
\[ U_1 = 0.01 J \]

### Step 3: Calculate the final energy with \( C_2 = 10μF \)
Given:
- \( C_2 = 10μF = 10 \times 10^{-6} F \)

Using the formula:
\[ U_2 = \frac{1}{2} C_2 V^2 \]
\[ U_2 = \frac{1}{2} \times (10 \times 10^{-6}) \times (100)^2 \]
\[ U_2 = \frac{1}{2} \times (10 \times 10^{-6}) \times 10000 \]
\[ U_2 = \frac{1}{2} \times 10 \times 10^{-2} \]
\[ U_2 = 5 \times 10^{-2} J \]
\[ U_2 = 0.05 J \]

### Step 4: Calculate the change in energy
The change in energy (\( \Delta U \)) is given by:
\[ \Delta U = U_2 - U_1 \]
\[ \Delta U = 0.05 J - 0.01 J \]
\[ \Delta U = 0.04 J \]

### Final Answer
The change in energy is:
\[ \Delta U = 0.04 J \]

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