Electrostatic energy stored in a capacitor of 12 pF capacitance connected to a 50 V battery is:

Options

A. `1 cdot 5xx 10^(-8) J`
B. `3xx 10^(-4) J`
C. `1 cdot 5xx10^(-10) J`
D. `9xx10^(-3) 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 energy stored in the capacitor,
- \( C \) is the capacitance,
- \( V \) is the voltage across the capacitor.

### Step 1: Identify the values of capacitance and voltage
Given:
- Capacitance \( C = 12 \, \text{pF} = 12 \times 10^{-12} \, \text{F} \)
- Voltage \( V = 50 \, \text{V} \)

### Step 2: Substitute the values into the formula
Now, substitute the values of \( C \) and \( V \) into the energy formula:

\[ U = \frac{1}{2} \times (12 \times 10^{-12} \, \text{F}) \times (50 \, \text{V})^2 \]

### Step 3: Calculate \( V^2 \)
First, calculate \( V^2 \):

\[ V^2 = 50^2 = 2500 \, \text{V}^2 \]

### Step 4: Substitute \( V^2 \) back into the formula
Now substitute \( V^2 \) back into the energy formula:

\[ U = \frac{1}{2} \times (12 \times 10^{-12}) \times 2500 \]

### Step 5: Calculate the energy
Now, calculate the energy:

\[ U = \frac{1}{2} \times 12 \times 2500 \times 10^{-12} \]

Calculating \( 12 \times 2500 \):

\[ 12 \times 2500 = 30000 \]

Now, substituting this back:

\[ U = \frac{1}{2} \times 30000 \times 10^{-12} = 15000 \times 10^{-12} \, \text{J} \]

### Step 6: Final result
Convert \( 15000 \times 10^{-12} \, \text{J} \) to a more convenient form:

\[ U = 15 \times 10^{-9} \, \text{J} = 15 \, \text{nJ} \]

Thus, the electrostatic energy stored in the capacitor is:

\[ \boxed{15 \, \text{nJ}} \]