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WAEC Physics 2020 Paper

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Question : 28

Total: 51

The magnitude of the force experienced by a charge of

1.6 \times 10^{-8} \mathrm{C}

in a uniform electric field of intensity

5 \times 105 \mathrm{NC}^{-1}

$3.2 \times 10^{-14} \mathrm{N}$
$8.0 \times 10^{-13} \mathrm{N}$
$8.0 \times 10^{-3} \mathrm{N}$
$3.1 \times 10^{13} \mathrm{N}$

Solution:

Concept:

The force on a charge placed in a uniform electric field is given by

F = qE

.

Explanation:

Here, charge

q = 1.6 \times 10^{-8} \ \mathrm{C}

and electric field intensity

E = 5 \times 10^{5} \ \mathrm{NC}^{-1}

.

Substitute these values into

F = qE

:

F = (1.6 \times 10^{-8})(5 \times 10^{5})

F = 8.0 \times 10^{-3} \ \mathrm{N}

So the correct option is C.

Answer:

8.0 \times 10^{-3} \ \mathrm{N}

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