Electrostatic enegy of `3.5xx10^(-4)J` is stored in a capacitor of 700 V. what is the charge on the capacitor?

Options

A. `5muC`
B. `4muC`
C. `1muC`
D. `8muC`

Explanation

To find the charge on the capacitor given the electrostatic energy and the voltage, we can use the formula for the energy stored in a capacitor:

\[ E = \frac{1}{2} Q V \]

Where:
- \( E \) is the energy stored in the capacitor (in joules),
- \( Q \) is the charge on the capacitor (in coulombs),
- \( V \) is the voltage across the capacitor (in volts).

### Step-by-step Solution:

1. **Identify the given values**:
- Electrostatic energy, \( E = 3.5 \times 10^{-4} \, \text{J} \)
- Voltage, \( V = 700 \, \text{V} \)

2. **Rearrange the formula to solve for charge \( Q \)**:
\[ E = \frac{1}{2} Q V \implies Q = \frac{2E}{V} \]

3. **Substitute the known values into the equation**:
\[ Q = \frac{2 \times (3.5 \times 10^{-4})}{700} \]

4. **Calculate the numerator**:
\[ 2 \times (3.5 \times 10^{-4}) = 7.0 \times 10^{-4} \]

5. **Now divide by the voltage**:
\[ Q = \frac{7.0 \times 10^{-4}}{700} \]

6. **Perform the division**:
\[ Q = 1.0 \times 10^{-6} \, \text{C} \]

7. **Convert to microcoulombs**:
\[ Q = 1.0 \, \mu\text{C} \]

### Final Answer:
The charge on the capacitor is \( Q = 1 \, \mu\text{C} \).

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