In the previous problem, suppose that another force in addition to the electrical force acts on the particle so that when it is released from rest, it moves to the right. After it has moved 5 cm, the additional force has done 9 × 10 -5 J of work and the particle has 4.5 × 10 -5 J of kinetic energy.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
[ – 4.5 × 10 -5 J
3 × 10 5 N/C
– 1.5 × 10 4 V]
Sol.
W ext + W elec = Δ K,
∴ 9 × 10 –5 + W elec = 4.5 × 10 –5
∴ W elec = – 4.5 × 10 –5 J
|W elec | = qEr = 4.5 × 10 –5 J
∴ E = 
= 3 × 10 5 N/C
V B – V A = – Er = – 3 × 10 5 × 5 × 10 –2 = – 1.5 × 10 4 V
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