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JAMB Physics 2023 Paper

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Question : 43

Total: 65

A charge of

4.6 \times 10^{-5}

C is placed in an electric field of intensity

3.2 \times 10^{4}

\mathrm{Vm}^{-1}

. What is the force acting on the electron?

1.5 N
3.7 N
4.2 N
2.5 N

Solution:

q= $4.6 \times 10^{-5}$ , C= $3.2 \times 10^{4}$ $\mathrm{Vm}^{-1}$ F= ?

F=qE

⇒

4.6 \times 10^{-5}

C ×

3.2 \times 10^{4}

; F= 1.5 N

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