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JAMB Physics 2024 Paper

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Question : 77

Total: 125

The charge of magnitude 1.6 x 10

^{-19}

C is placed in a uniform electric field of intensity 1200Vm

^{-1}

. Calculate its acceleration, if the mass of the charge is 9.1 x 10

^{-31}

kg

2.11 x 10 $^{14}$ ms $^{-2}$
2.11 x 10 $^{13}$ ms $^{-2}$
1.20 x 10 $^{14}$ ms $^{-2}$
1.20 x 10 $^{13}$ ms $^{-2}$

Solution:

Concept:

The electric field exerts a force on the charge, and Newton’s second law gives the resulting acceleration.

Explanation:

Electric field intensity is defined as

E = \frac{F}{q}

, so the force is

F = Eq

.

Given:

E = 1200 \, \mathrm{Vm^{-1}}

,

q = 1.6 \times 10^{-19} \, \mathrm{C}

, and

m = 9.1 \times 10^{-31} \, \mathrm{kg}

.

Calculate the force:

F = Eq = 1200 \times 1.6 \times 10^{-19} = 1.92 \times 10^{-16} \, \mathrm{N}

Use Newton’s second law,

F = ma

, to find acceleration:

a = \frac{F}{m} = \frac{1.92 \times 10^{-16}}{9.1 \times 10^{-31}}

a = 2.11 \times 10^{14} \, \mathrm{ms^{-2}}

Answer:

A.

2.11 \times 10^{14} \, \mathrm{ms^{-2}}

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