Question

Two 2.3-cm-diameter-disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.6×105 V/m .

Part A

What is the voltage across the capacitor?

Part B

How much charge is on each disk?

Part C

An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.3×10⁷ m/s . What was the electron's speed as it left the negative plate?

Answer 1

here,

diameter of each disk , d = 2.3 cm = 0.023 m

radius , r = d/2 = 0.0115 m

spacing between them , s = 2 mm = 0.002 m

electric feild , E = 4.6 * 10^5 V/m

a)

the voltage across the capacitor , V = E * s

V = 4.6 * 10^5 * 0.002 = 920 V

b)

the charge on disk , Q = C * V

Q = (area * e0 /s) * V

Q = (pi * 0.0115^2 * 8.85 * 10^-12 /0.002) * 920

Q = 1.69 * 10^-9 C

c)

u = 2.3 * 10^7 m/s

let the final speed be v

using work energy theorm

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

920 * 1.6 * 10^-19 = 0.5 * 9.1 * 10^-31 * ( v^2 - ( 2.3 * 10^7)^2)

solving for v

v= 2.92 * 10^7 m/s

the final speed is 2.92 * 10^7 m/s