A parallel-plate capacitor consists of two square plates that measure 90 cm by 9

Question

A parallel-plate capacitor consists of two square plates that measure 90 cm by 90 cm. An electron is launched at a 45° angle and a speed of 5x10^6 m/s from the center of the positive plate towards the negative plate. The electron hits the positive plate 4 cm away from where it was launched.

A) Prepare a graph. and Compute the electric field strength inside the capacitor.

B) Compute the charge Q on the positive plate.

C) Compute the smallest possible spacing between the two plates that allows the electron to move as described.

Explanation / Answer

Part a)

Electron will undergo a projectile motion

Range = (v2/a) sin 90 = 0.04

a = v2/0.04 = (5*106)2/(0.04)

a = 6.25*1014 m/s2;

Force on electron = (9.11*10-31)*6.25*1014

Force = 5.7*10-16 N

F= eE

E = F/e = 5.7*10-16/(1.610-19) = 3.6*103 N/C

So, electric field strength inside the capacitor = 3.6*103 N/C

Part c)

v2 = 2aH

H = v2/2a

H = 25*1012/(2*6.25*1014)

= 2*10-2 m

= 20 mm or 2 cm

Maximum spacing = 2 cm

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