Question

Two 2.2-cm-diameter-disks spaced 1.9 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.8×105 V/m .

a) How much charge is on each disk?

b) An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.0×107 m/s . What was the electron's speed as it left the negative plate?

Answer 1

Given,

D = 2.2 cm => r = 1.1 cm = 0.011 m ; d = 1.9 mm ; E = 4.8 x 10^5 V/m

a)We know that

E = V/d => V = Ed

V = 4.8 x 10^5 x 1.9 x 10^-3 = 912 V

C = e0 A/d

A = pi r^2

C = 8.85 x 10^-12 x 3.14 x 0.011^2/1.9 x 10^-3 = 1.76 x 10^-12 F

Q = CV = 1.76 x 10^-12 x 912 = 1.605 x 10^-9 C

Hence, Q = 1.605 x 10^-9 C

b)v= 2 x 10^7

We know that, F = qE = ma

a = qE/m

a = 1.6 x 10^-19 x 4.8 x 10^5/(9.1 x 10^-31) = 8.44 x 10^16 m/s^2

from eqn of motion

v^2 = u^2 + 2 a S

u^2 = v^2 - 2 a S

u^2 = (2 x 10^7)^2 - 2 x 8.44 x 10^16 x 1.9 x 10^-3 = 0.793 x 10^14

u = 8.9 x 10^6 m/s