Question

A parallel plate capacitor is formed with two plates separated by 5.00 mm as shown in Figure 1. Each plate is a 10.0 cm X 10.0 cm square. We do not know the charge on the plates. An electron beam is shot in from one edge of the capacitor. It enters the capacitor very close (call it 0 mm) from the top plate and travelling parallel to the plates. The electrons in the beam are moving at 1.50 × 107 m/s as they enter the capacitor (this may seem extremely fast, but it is easy to make electrons go this fast). They exit the other side of the capacitor just missing the edge of the bottom plate.

(a) Which plate is the positive one? Explain.

(b) From the information given determine the acceleration of the electrons.

(c) Find the strength of the electric field inside the capacitor.

(d) Determine how much charge is on each plate of the capacitor.

5 mm 10 cm An electron beam fired into a parallel plate capacitor.

Answer 1

(a) As the e beam is deflected downward, the electric field must be acting in upward direction from the down plate to the upper one. So, lower plate must be positively charged.

(b) Initially, the beam has only horizontal velocity which remain constant through the length of the plate, l. So, using, s = vt, we have, time taken to travel the length of the plates,

t = s/v = l/Vx

Along vertical, initial velocity, u = 0

But the beam will experience a force due to which it is deflected downward, the acceleration, a ,for which can be found by using kinematic equation, s = ut + 1/2 at^2, where, s = vertical displacement of the beam, y = 5 mm = 0.005 m, u = 0, t = l/Vx = 0.1/1.5*10^7 = 6.7*10^-9 s, hence,

y = 0 + 1/2 at^2 => a = 2y/t^2

   = 2*0.005/(6.7*10^-9)^2

Thus, acceleration, a = 1.49*10^15 m/s^2

(c) Again, a = eE/m, where, e = 1.67*10^-19 C,

m = mass of electron = 9.1*10^-31 kg, so,

Electric field, E = ma/e

= 9.1*10^-31*1.49*10^15/1.67*10^-19

= 8.12*10^3 N/C

(d) Now, potential between plates, V = Ed

Where, d = separation between plates = 0.005 m

and, capacitance, C = $\dpi{200} \epsilon$ oA/d

Hence, charge on the capacitor plates

Q = CV = $\dpi{200} \epsilon$ oA/d * Ed = $\dpi{200} \epsilon$ oAE

Therefore, Q = 8.85*10^-12*0.01*8.12*10^3

= 7.186*10^-10 C