A negatively charged particle is released from rest at point A and accelerates u
ID: 1538067 • Letter: A
Question
A negatively charged particle is released from rest at point A and accelerates until it reaches point B. The mass and charge of this particle are 4.0 mg and -20. µC, respectively. Only the gravitational force and the electrostatic force act on this particle, which moves in a horizontal straight line without rotating. The electric potential at B is 36 V greater than at A. a) How much electric potential energy does it lose? b) How much kinetic energy does it gain? c) What is the particle’s speed when it reaches point B?
Explanation / Answer
(A)
electric potential energy lost = dV*q
dV = potential energy = 36 V
q = charge = -20*10^-6 C
electric potential energy lost = 36*20*10^-6 = 0.00072 J
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(b)
work energy relation
dKE = dU = 0.00072 J
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(b)
iniital KE k1 = 0
final KE K2 = (1/2)*m*v^2
(1/2)*m*v^2 - 0= 0.00072
m = mass = 4*10^-6 kg
(1/2)*4*10^-6*v^2 = 0.00072
v = 18.97 m/s <<<<----------answer
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