Question

A \(10\,pF\) capacitor is connected to a \(24\,V\) battery. The electrostatic energy stored in the capacitor is:

• A. \(11.52\times10^{-9}\,J\)
• B. \(1.2\times10^{-9}\,J\)
• C. \(5.76\times10^{-9}\,J\)
• D. \(2.88\times10^{-9}\,J\)

Hint:

For capacitor energy: \[ U=\frac{1}{2}CV^2 \] A quick shortcut: \[ U(\text{J}) = 0.5\times C(F)\times V^2 \] Always convert pF into farads first: \[ 1\,pF=10^{-12}F \]

Answer 1

The Correct Option is D

Step 1: Recall the energy stored in a capacitor. \[ U=\frac{1}{2}CV^2 \] Given: \[ C=10\,pF=10\times10^{-12}F \] \[ V=24V \]

Step 2: Substitute the values. \[ U = \frac{1}{2} \left(10\times10^{-12}\right) (24)^2 \] \[ = \frac{1}{2} \left(10\times10^{-12}\right) (576) \] \[ = 5\times10^{-12}\times576 \] \[ = 2880\times10^{-12} \] \[ = 2.88\times10^{-9}\,J \]

Step 3: Identify the correct option. \[ \boxed{U=2.88\times10^{-9}\,J} \]