Question

Problem 5 When an empty parallel-plate capacitor with a cross sectional area of 3x10m2 is connected across 50V, 5nC of charge flows to each plate. (6 points each) (a) What is the separation of the plates? (b) How much energy is stored in this capacitor? (c) If an electron is relcased from rest at the negative plate then what is its speed when it hits the positive plate?

Answer 1

here,

5)

cross sectional area , A = 3 * 10^-4 m^2

potential difference between the plates , V = 50 V

charge , Q = 5 nC = 5 * 10^-9 C

a)

let the sepration between plates be d

capacitance , C = Q/V = area * e0 /d

5 * 10^-9 /50 = 3 * 10^-4 * 8.85 * 10^-12 /d

solving for d

d = 2.66 * 10^-5 m

b)

the energy stored , E = 0.5 * Q * V

E = 0.5 * (5 * 10^-9 ) * ( 50) J

E = 1.25 * 10^-7 J

c)

let the speed at positive plate be v

using work energy theorm

work done by potential = kinetic energy gained

V * e = 0.5 * m * ( v^2 - 0)

50 * 1.6 * 10^-19 = 0.5 * 9.1 * 10^-31 * v^2

solving for v

v = 4.2 * 10^6 m/s