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JAMB Physics 2013 Paper

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Question : 29

Total: 43

Calculate the force acting on an electron of charge 1.5 x 10^-19C placed in an electric field of intensity 10⁵Vm^-1

1.5x10^-11N
1.5x10^-12N
1.5x10^-13N
1.5x10^-14N

Solution:

Concept:

The electric force is found by multiplying the charge by the electric field intensity.

Explanation:

Use the relation

F = qE

.

Here,

q = 1.5 \times 10^{-19}\,\mathrm{C}

and

E = 10^{5}\,\mathrm{Vm}^{-1}

.

Substitute:

F = (1.5 \times 10^{-19})(10^{5})

.

Add the exponents of 10:

F = 1.5 \times 10^{-19+5} = 1.5 \times 10^{-14}\,\mathrm{N}

.

Answer:

The correct option is D:

1.5 \times 10^{-14}\,\mathrm{N}

.

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