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: Understanding the Concept:
This problem follows the exact same underlying physics principles of capacitor energy storage as the previous question. The amount of mechanical-electrical energy stored varies linearly with capacitance and scales quadratically with the applied electrical potential difference.
Step 2: Key Formula or Approach:
We apply the standard potential energy equation: $$ U = \frac{1}{2} C V^2 $$ Let's convert our parameters into base metric units: - Capacitance ($C$): $10 \text{ pF} = 10 \times 10^{-12} \text{ F} = 10^{-11} \text{ F}$ - Potential difference ($V$): $24 \text{ V}$
Step 3: Detailed Explanation:
Let's substitute our values directly into the energy formula: $$ U = \frac{1}{2} \times \left(10 \times 10^{-12} \text{ F}\right) \times (24 \text{ V})^2 $$ First, compute the square of the voltage multiplier: $$ (24)^2 = 576 $$ Substitute this product back into our tracking equation and compute the division: $$ U = \frac{1}{2} \times 10 \times 10^{-12} \times 576 $$ $$ U = 5 \times 576 \times 10^{-12} $$ $$ U = 2880 \times 10^{-12} \text{ J} $$ Let's adjust the decimal indices to line up with the standard $10^{-9}\text{ J}$ choice options: $$ U = 2.88 \times 10^3 \times 10^{-12} \text{ J} $$ $$ U = 2.88 \times 10^{3 - 12} \text{ J} = 2.88 \times 10^{-9} \text{ J} $$ This calculated outcome matches option (D).
Step 4: Final Answer:
The electrostatic energy stored in the capacitor is 2.88 $\times$ 10$^{-9}$ J.