A charge of 4.6×10−5C is placed in an elec...
A charge of 4.6×10−5C is placed in an electric field of intensity 3.2×104 Vm−1. What is the force acting on the electron?
1.5 N
3.7 N
4.2 N
2.5 N
Correct answer is A
q=4.6×10−5, C= 3.2×104 Vm−1 F= ?
F=qE
⇒ 4.6×10−5C × 3.2×104
; F= 1.5 N
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