A positive charge of 3 × 10–8C is in a uniform electric field of field strength 300 000V m–1. How much work must be done on the charge in order to move it a distance of 4.0 mm in the opposite direction to the direction of the field?​

Answers (1)

Force F=qE

F=2*10-8*300000=6*10-3N

W=force*displacement=-6*10-3*4*10-3=24*10-6J

Negative sign indicates that work done is against the force

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