A particle of charge = 50 µC moves in a region where the only force on it is an electric force. As the particle moves 25 cm, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference acting on the partice
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Ответ:
+60 J
Explanation:
Work done by the field on the charge is given by W = qEd where q = charge = +4.0 C, E= electric field strength = 30 N/C and d = distance moved by charge = 0.50 m
So, substituting the values of the variables into the equation, we have
W = qEd
= +4.0 C × 30 N/C × 0.50 m
= +60 Nm
= +60 J