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WAEC Physics 2014 Paper

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Question : 40

Total: 48

Two identical charges situated 20cm apart in vacuum repel each other with a force of 1.0N.Calculate the value of each other with a force of 1.0N. Calculate the value of each charge. [(14

(\pi \varepsilon_o)^{-1}

= 9 x 10⁹Nm²C^-2]

21 $\mu\mathrm{C}$
3.6 $\mu\mathrm{C}$
4.0 $\mu\mathrm{C}$
4.4 $\mu\mathrm{C}$

Solution:

Concept:

Coulomb's law gives the repulsive force between two identical point charges.

Use

F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2}

where the constant

k = \frac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{Nm}^2\text{C}^{-2}

.

Explanation:

Both charges are equal, so let each charge be

q

.

Convert distance:

r = 20\ \text{cm} = 0.20\ \text{m}

.

Thus,

F = k\frac{q^2}{r^2}

.

Substitute

F = 1.0\ \text{N}

and

r = 0.20\ \text{m}

:

1.0 = (9\times10^9)\frac{q^2}{(0.20)^2}

q^2 = \frac{1.0 \times (0.20)^2}{9\times10^9} = \frac{0.04}{9\times10^9}

q^2 = 4.44\times10^{-12}

q = \sqrt{4.44\times10^{-12}} = 2.11\times10^{-6}\ \text{C}

Since

1\ \mu\text{C} = 10^{-6}\ \text{C}

, we have

q = 2.11\ \mu\text{C}

.

Answer:

The value of each charge is

2.11\times10^{-6}\ \text{C}

, which is approximately

2.1\ \mu\text{C}

.

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