A 900 pF capacitor is charged by a 100 V battery. The electrostatic energy stored by the capacitor is

Options

A. `9xx10^(5)J`
B. `4.5xx10^(-6)J`
C. `2.2xx10^(-6)J`
D. `9xx10^(-10)J`

Answer

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,
- \( C \) is the capacitance in Farads,
- \( V \) is the voltage in Volts.

### Step-by-step Solution:

**Step 1: Convert the capacitance from picoFarads to Farads.**

Given:
- Capacitance \( C = 900 \, \text{pF} \)

To convert picoFarads to Farads:
\[ C = 900 \, \text{pF} = 900 \times 10^{-12} \, \text{F} \]

**Step 2: Identify the voltage.**

Given:
- Voltage \( V = 100 \, \text{V} \)

**Step 3: Substitute the values into the energy formula.**

Using the formula:
\[ U = \frac{1}{2} C V^2 \]

Substituting the values:
\[ U = \frac{1}{2} \times (900 \times 10^{-12}) \times (100)^2 \]

**Step 4: Calculate \( V^2 \).**

Calculating \( V^2 \):
\[ V^2 = 100^2 = 10000 \]

**Step 5: Substitute \( V^2 \) back into the equation.**

Now substituting \( V^2 \) into the equation:
\[ U = \frac{1}{2} \times (900 \times 10^{-12}) \times 10000 \]

**Step 6: Simplify the expression.**

Calculating:
\[ U = \frac{1}{2} \times (900 \times 10^{-12}) \times (10^4) \]
\[ U = \frac{1}{2} \times (900 \times 10^{-8}) \, \text{J} \]
\[ U = 450 \times 10^{-8} \, \text{J} \]

**Step 7: Convert to standard form.**

\[ U = 4.5 \times 10^{-6} \, \text{J} \]

### Final Answer:
The electrostatic energy stored by the capacitor is \( 4.5 \times 10^{-6} \, \text{J} \) or \( 4.5 \, \mu\text{J} \).