Imagine moving a proton from the negative plate to the positive plate of a paral
ID: 1581561 • Letter: I
Question
Imagine moving a proton from the negative plate to the positive plate of a parallel plate
arrangement. The plates are 2.00 cm apart, and the field is uniform with a magnitude of
3000 N/C. .
a) What is the voltage between the plates?
Answer: 60V
b) What is the change in electric potential energy of the proton? Your answer must have the
correct sign.
Answer:+9.6x10-18 J
c) If the proton is released from rest at the positive plate, what speed will it have just before it
hits the negative plate?
Answer: 1.1x10^5 m/s
Explanation / Answer
a) Voltage between the plates, V = E*d
= 3000*2*10^-2
= 60 V
b) change in potential energy = q*(V2 - V1)
= 1.6*10^-19*60
= +9.6*10^-18 J
c) Workdone on the proton = gain in kinetic energy
q*V = (1/2)*m*v^2
==> v = sqrt(2*q*V/m)
= sqrt(2*1.6*10^-19*60/(1.67*10^-27))
= 1.1*10^5 m/s