Question

A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is

• A. 15
• B. 7.5
• C. 0.3
• D. 150

Answer 1

The Correct Option is A

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

\(U = \frac{1}{2} C V^2\) 

where:

• \(U\) is the electrostatic energy stored in the capacitor,
• \(C\) is the capacitance of the capacitor,
• \(V\) is the voltage across the capacitor.

Given values:

• \(C = 12 \, \text{pF} = 12 \times 10^{-12} \, \text{F}\)
• \(V = 50 \, \text{V}\)

Now, plugging in the values into the formula:

\(U = \frac{1}{2} \times 12 \times 10^{-12} \times (50)^2\)

Calculating further:

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

\(U = 6 \times 10^{-12} \times 2500\)

\(U = 15000 \times 10^{-12} \, \text{J}\)

\(U = 15 \times 10^{-9} \, \text{J} = 15 \, \text{nJ}\)

Therefore, the electrostatic energy stored in the capacitor is \(15 \, \text{nJ}\).

The correct answer is 15 nJ.