Question

A 15pF capacitor is connected to 60V battery. How much electrostatic energy is stored?

Answer 1

ENERGY STORED IS 27 × 10^(-9) Joule.

Given:

• Capacitance = 15 pF
• Potential difference provided by battery = 60 volts.

To find:

Electrostatic energy stored in capacitor?

Calculation:

The general expression for ELECTROSTATIC energy stored in the capacitor is given as:

$E = \dfrac{1}{2} \times C \times {V}^{2}$

• E is energy
• C is capacitance
• V is potential difference

$\implies E = \dfrac{1}{2} \times (15 \times {10}^{ - 12} ) \times {(60)}^{2}$

$\implies E = \dfrac{1}{2} \times 15 \times {10}^{ - 12} \times 36 \times {10}^{2}$

$\implies E = 270 \times {10}^{ - 10}$

$\implies E = 27 \times {10}^{ - 9} \: joule$

So, energy stored is 27 × 10^(-9) Joule