A point charge of 7.57*10^-6C and mass 68.7*10^-6 kg is uniform electric feild of 474 N/C i Released from rest, the charges is accelerated by the field obtaining a velocity of 74.6 m/s i. over what distance did the change its speed?
A point charge of 7.57*10^-6C and mass 68.7*10^-6 kg is uniform electric feild of 474 N/C i Released from rest, the charges is accelerated by the field obtaining a velocity of 74.6 m/s i. over what distance did the change its speed?
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Question
A point charge of 7.57*10^-6C and mass 68.7*10^-6 kg is uniform electric feild of 474 N/C i Released from rest, the charges is accelerated by the field obtaining a velocity of 74.6 m/s i. over what distance did the change its speed?
Expert Solution
Step 1
Given:
charge, q = 7.57 * 10-6 C
mass, m = 68.7 * 10-6 kg
electric field, E = 474 N/C
initial velocity, u = 0 m/s
final velocity, v = 74.6 m/s
Step 2
Force acting on the point charge in the electric field is given as,
F = E q
F = 474 N/C * 7.57 * 10-6 C
F = 3.588 * 10-3 N
From Newton's second law, we have : F = ma
Therefore, the acceleration of the point charge, a = F/m
a = 3.588 * 10-3 N / 68.7 * 10-6 kg
a = 52.23 m/s2
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