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?
A
1.5 N
B
3.7 N
C
4.2 N
D
2.5 N
Verified Correct Answer: A.
Physical Principles & Numerical Solution
q=\(4.6 \times 10^{-5}\), C= \(3.2 \times 10^4\) \(Vm^{-1}\) F= ?
F=qE
⇒ \(4.6 \times 10^{-5}\)C \times \(3.2 \times 10^4\)
; F= 1.5 N
Source: WAEC / JAMB / NECO Classroom Archive (Record #68565)