Question

A parallel plate capacitor consists of two square plates, each 2 cm on a side, with a separation of 2 mm. The magnitude of the charge on each plate is 4 nC. An electron is released from rest at the negative plate and accelerates towards the positive plate. What is the magnitude of its velocity when it reaches the positive plate?

Answer 1

Step 1: Find capacitance of the capacitor:

C = e0*A/d

A = Area of plate = 2 cm*2 cm = 4 cm = 0.04 m

d = separation distance = 2 mm = 2*10^-3 m

So

C = 8.854*10^-12*4*10^-4/(2*10^-3)

C = 1.77*10^-12 F

Step 2: Find potential difference between plates

Since Q = C*dV

dV = Q/C

dV = 4*10^-9/(1.77*10^-12) = 2260 V

Step 3:

Using energy conservation find speed of electron released:

KEi + PEi = KEf + PEf

KEi = 0, since electron released from rest

KEf = (1/2)*m*v^2, where v is speed of electron at positive plate

PEf - PEi = dU = q*dV

So,

KEf = PEi - PEf

KEf = -dU = -q*dV

q = charge on electron = -1.6*10^-19 C

So,

(1/2)*m*v^2 = -q*dV

v = sqrt (-2*q*dV/m)

Using known values:

v = sqrt (-2*(-1.6*10^-19)*2260/(9.1*10^-31))

v = 2.82*10^7 m/s = speed of electron at positive plate

Let me know if you've any query.