A capacitor of capacitance 700 pF is charged by 100 V battery. The electrostatic energy stored by the capacitor is

Options

A. `2.5xx10^(-8)J`
B. `3.5xx10^(-6)J`
C. `2.5xx10^(-4)J`
D. `3.5xx10^(-4)J`

Explanation

To find the electrostatic energy stored in a capacitor, we can use the formula:

\[ U = \frac{1}{2} C V^2 \]

where:
- \( U \) is the electrostatic energy stored (in joules),
- \( C \) is the capacitance (in farads),
- \( V \) is the voltage (in volts).

### Step-by-step solution:

1. **Identify the given values**:
- Capacitance \( C = 700 \, \text{pF} = 700 \times 10^{-12} \, \text{F} \)
- Voltage \( V = 100 \, \text{V} \)

2. **Substitute the values into the formula**:
\[ U = \frac{1}{2} C V^2 = \frac{1}{2} \times (700 \times 10^{-12}) \times (100)^2 \]

3. **Calculate \( V^2 \)**:
\[ V^2 = 100^2 = 10000 \]

4. **Substitute \( V^2 \) back into the equation**:
\[ U = \frac{1}{2} \times (700 \times 10^{-12}) \times 10000 \]

5. **Simplify the expression**:
\[ U = \frac{1}{2} \times (700 \times 10^{-12}) \times (10^4) = \frac{1}{2} \times (700 \times 10^{-8}) = 350 \times 10^{-8} \, \text{J} \]

6. **Convert to standard form**:
\[ U = 3.5 \times 10^{-6} \, \text{J} \]

### Final Answer:
The electrostatic energy stored by the capacitor is:
\[ U = 3.5 \times 10^{-6} \, \text{J} \]