A parallel plate capacitor is constructed from two plates that are 4 cm by 3 cm,
ID: 1559530 • Letter: A
Question
A parallel plate capacitor is constructed from two plates that are 4 cm by 3 cm, separated by a 1 cm gap. The plates are charged to +/- 30 nC. a. What is the magnitude of the [electric field between the plates? A proton and an electron are both placed halfway between the plates (see figure), then released. (Ignore any effects they have on each other.) b. What is the final speed of the proton when it hits the negative plate? c. What is the final speed of the electron when it hits the positive plate?Explanation / Answer
given
Q = 30 nC
A = 4*3
= 12 cm^2
= 12*10^-4 m^2
d = 1 cm = 0.01 m
a) E = Q/(A*epsilon)
= 30*10^-9/(12*10^-4*8.854*10^-12)
= 2.82*10^6 N/C
b) Workdone on proton = gain in kinetic energy
F*(d/2) = (1/2)*mp*v^2
q*E*(d/2) = (1/2)*mp*v^2
==> v = sqrt(q*E*d/mp)
= sqrt(1.6*10^-19*2.82*10^6*0.01/(1.67*10^-27))
= 1.64*10^6 m/s
c) Workdone on electron = gain in kinetic energy
F*(d/2) = (1/2)*me*v^2
q*E*(d/2) = (1/2)*me*v^2
==> v = sqrt(q*E*d/me)
= sqrt(1.6*10^-19*2.82*10^6*0.01/(9.11*10^-31))
= 7.04*10^6 m/s