Question

Problem 2 – The plates of a parallel-plate capacitor are separated by a distance d = 0.2 m. There is vacuum between the plates. The voltage difference between the plates is 150 V. The capacitance of the plates is 3 μF.

a) (6 pts) Find the magnitude of the electric field between the plates (ignoring edge effects).

b) (12 pts) An alpha particle, which is doubly ionized helium, He2+ (charge = 2e where e is the elementary charge, mass = 4.7 × 10-27 kg) is released from rest at the positively charged plate. Find the speed of the alpha particle as it hits the negatively charged plate.

c) (6 pts) Find the magnitude of the total charge on one of the plates of the capacitor.

d) (6 pts) Find the total energy stored in the capacitor.

Answer 1

Part A.

Electric field between the plates of capacitor is given by:

E = V/d

Given that, V = Potential difference = 150 V

d = separation distance between plates = d = 0.2 m

So,

E = 150/0.2

E = 750 N/C

Part B.

Using Force balance on alpha particle moving between plates:

F_net = Fe

m*a = q*E

a = acceleration of alpha particle = q*E/m

a = 2*1.6*10^-19*750/(4.7*10^-27) = 5.11*10^10 m/s^2

Now using 3rd kinematic equation:

V^2 = U^2 + 2*a*d

U = Initial speed of alpha particle = 0 m/s, since released from rest

V = sqrt (0^2 + 2*5.11*10^10*0.2)

V = 1.43*10^5 m/s = Speed of alpha particles when it reaches on negatively charged plate

Part C.

Q = C*V

C = Capacitance of capacitor = 3*10^-6 F

So,

Q = 3*10^-6*150

Q = 4.5*10^-4 C = Charge on one of the plates

Part D.

Energy stored in capacitor is given by:

U = (1/2)*C*V^2

U = (1/2)*3*10^-6*150^2

U = 337.5*10^-4 J

U = 3.375*10^-2 J = energy stored in capacitor

Let me know if you've any query.