Two 2.0-cm-diameter-disks spaced 1.6 mm apart form a parallel-plate capacitor. T

Question

Two 2.0-cm-diameter-disks spaced 1.6 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.0×105 V/m .

Part A

What is the voltage across the capacitor?

Express your answer with the appropriate units.

Part B

How much charge is on each disk?

Enter your answers numerically separated by a comma.

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?

Express your answer with the appropriate units.

Explanation / Answer

(A)

d = 1.6 mm = 0.0016 m

E = 4*10^5 N/C

so, voltage across capacitor is,

V = E*d = 640 V

(B)

charge q = C*V

and C = e0*A / d

C = 8.85*10^(-12)*pi*(0.01)^2 / 0.0016

C = 1.73*10^(-12) C

hence q = 1.73*10^(-12)*640

q = 1.11 nC

(C)

qe*E = m*a

a = q*E / m = 1.16*10^(-19)*4*10^5 / 9.1*10^(-31)

a = 0.703*10^17 m/s^2

let initial speed of electron = u

apply equation of motion,

v^2 - u^2 = 2*a*d

u^2 = (2.3*10^7)^2 - 2*0.703*10^17*0.0016

u = 1.74*10^7 m/s

answer

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