A parallel plate capacitor with a capacitance of 12\text{ }\mu\text{F} is fully charged by… — JAMB UTME

A parallel plate capacitor with a capacitance of $12\text{ }\mu\text{F}$ is fully charged by connecting it across a $50\text{ V}$ direct current power source. What is the total electrostatic energy stored in the capacitor in millijoules ( $\text{mJ}$ )?

Answer: 15 mJ

Answer

The total electrostatic energy stored in the capacitor is

15\text{ mJ}

The energy stored in a capacitor is given by

E = \frac{1}{2}CV^2

. Substituting

C = 12 \times 10^{-6}\text{ F}

and

V = 50\text{ V}

yields

E = \frac{1}{2} \times 12 \times 10^{-6} \times 2500 = 0.015\text{ J} = 15\text{ mJ}

Step-by-Step Solution

Convert given values to standard SI units.

C = 12 \times 10^{-6}\text{ F}

V = 50\text{ V}

Calculating in SI units ensures the resultant energy is in Joules.

Apply the formula for energy stored in a charged capacitor.

E = \frac{1}{2} C V^2

Work done during charging is stored as electrostatic potential energy in the electric field between the plates.

Substitute the values and convert Joules to millijoules.

E = \frac{1}{2} \times (12 \times 10^{-6}) \times 2500 = 0.015\text{ J} = 15\text{ mJ}

Multiplying Joules by

10^3

converts the result into millijoules.

Key Concept

Energy stored in a capacitor (

E = \frac{1}{2}CV^2

)

Question

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Step-by-Step Solution

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