Test Index

JAMB Physics 2023 Paper

© africaexams.com

Question : 20

Total: 65

A parallel plate capacitor separated by an air gap is made of

0.8\,\mathrm{m}^2

tin plates and 20 mm apart. It is connected to 120 V battery. What is the charge on each plate?

\varepsilon_o

8.85\times10^{-12}\,\mathrm{F}\,\mathrm{m}^{-1}

3.54nC
42.5nC
35.4nC
4.25nC

Solution:

Concept:

A parallel-plate capacitor stores charge

Q

on each plate when connected to a voltage

V

, using the relation

Q = CV

.

Explanation:

Given:

A = 0.8\,\mathrm{m}^2

,

d = 20\,\mathrm{mm} = 0.02\,\mathrm{m}

,

V = 120\,\mathrm{V}

,

\varepsilon_o = 8.85 \times 10^{-12}\,\mathrm{F\,m}^{-1}

.

Capacitance is given by

C = \dfrac{\varepsilon_o A}{d}

.

Substitute the values:

C = \dfrac{8.85 \times 10^{-12} \times 0.8}{0.02}

.

This gives

C = 3.54 \times 10^{-10}\,\mathrm{F}

.

Now,

Q = CV = (3.54 \times 10^{-10})(120)

.

Thus,

Q = 4.248 \times 10^{-8}\,\mathrm{C}

.

Converting to nanocoulombs:

Q = 42.48 \times 10^{-9}\,\mathrm{C} \approx 42.5\,\mathrm{nC}

.

Answer:

The charge on each plate is

42.5\,\mathrm{nC}

, which matches option B.

© africaexams.com

Go to Question:

Prev Question

Next Question