In: Physics
A point charge 4.80 μC is held fixed at the origin. A
second point charge 1.40 μC with mass of
2.80×10−4 kg is placed on the x axis, 0.260 m
from the origin.
Part A: What is the electric potential energy U of the
pair of charges? (Take U to be zero when the charges have
infinite separation.) (Answer in Joules)
Part B: The second point charge is released from rest. What is its
speed when its distance from the origin is 0.500 m . (Answer in
m/s)
Part C: What is its speed when its distance from the origin is 5.00m. (Answer in m/s).
Part D: What is its speed when its distance from the origin is 50.0m. (Answer in m/s).
Part A -
The expression for the electric potential energy = U1 =
kQq/r1
So, the electric potential energy=U1
=(9*19^9)[4.80*10^-6][1.40*10^-6] / 0.260 = 0.232 J
Part -B -
The second point charge is released from rest. What is its speed
when its distance from the origin is 0.500 m ?
The electric potential energy=U2=kQq/r2
=(9*19^9)[4.80*10^-6][1.40*10^-6] / 0.50 = 0.1210 J
Kinetic energy =U1 - U2= 0.232 - 0.121 = 0.111 J
Now, velocity = v1 = sq rt 2*kinetic energy / mass = sq rt [(2 *
0.111)/ (2.8*10^-4)] = 28.16 m/s
Part C -
The second point charge is released from rest. What is its speed
when its distance from the origin is 5.0 m?
The electric potential energy=U3=kQq/r3
=(9*19^9)[4.80*10^-6][1.40*10^-6] / 5.0 = 0.0121 J
Kinetic energy =U1 - U3= 0.232 - 0.0121 = 0.22 J
Therefore, velocity = v2 = sq rt 2*kinetic energy / mass = sq rt
(2* 0.22 / 2.8*10^-4) = 39.6 m/s
Part - D
The second point charge is released from rest. What is its speed
when its distance from the origin is 50.0 m?
The electric potential energy=U4=kQq/r4
=(9*19^9)[4.80*10^-6][1.40*10^-6]/ 50.0=0.00121 J
Kinetic energy =U1 - U4 = 0.232 - 0.00121 = 0.231 J
Velocity = v3 = sq rt 2*kinetic energy / mass = sq rt (2* 0.231 /
2.8*10^-4) = 40.60 m/s