A charge of \(4.6 × 10^{-5}\)C is placed in an electric field of intensity \(3.2 × 10^4\) \(Vm^{-1}\). What is the force acting on the electron?
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Correct Answer: Option A
Explanation:
q=\(4.6 × 10^{-5}\), C= \(3.2 × 10^4\) \(Vm^{-1}\) F= ?
F=qE
⇒ \(4.6 × 10^{-5}\)C × \(3.2 × 10^4\)
; F= 1.5 N
q=\(4.6 × 10^{-5}\), C= \(3.2 × 10^4\) \(Vm^{-1}\) F= ?
F=qE
⇒ \(4.6 × 10^{-5}\)C × \(3.2 × 10^4\)
; F= 1.5 N