Question

Find the energy stored in a capacitance of 10 μF when it is charged to a potential difference of 2 volts.

Hint:

The energy stored in a capacitor is proportional to the square of the potential difference across it.

Answer 1

Step 1: Formula for Energy Stored in a Capacitor.
The energy \( E \) stored in a capacitor is given by the formula: \[ E = \frac{1}{2} C V^2 \] Where:
- \( C = 10 \, \mu\text{F} = 10 \times 10^{-6} \, \text{F} \) is the capacitance,
- \( V = 2 \, \text{V} \) is the potential difference.
Step 2: Calculation.
Substituting the values into the formula: \[ E = \frac{1}{2} \times 10 \times 10^{-6} \times (2)^2 = \frac{1}{2} \times 10 \times 10^{-6} \times 4 = 20 \times 10^{-6} \, \text{J} = 0.01 \, \text{J} \]
Final Answer:
The energy stored in the capacitor is \( \boxed{0.01 \, \text{J}} \).