Question

Two 2.3-cm-diameter-disks spaced 1.6 mm apart form a parallel-plate capacitor. The electric field between the disks...

Two 2.3-cm-diameter-disks spaced 1.6 mm apart form a parallel-plate capacitor. The electric field between the disks is 5.0×105 V/m . Part A Part complete What is the voltage across the capacitor? Express your answer with the appropriate units. V = 800 V Previous Answers Correct Part B How much charge is on each disk? Enter your answers numerically separated by a comma. Express your answers in coulombs. q1, q2 = 1.836⋅10−9,1.836⋅10−9 C Previous AnswersRequest Answer Incorrect; Try Again; 5 attempts remaining Part C An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.4×107 m/s . What was the electron's speed as it left the negative plate? Express your answer with the appropriate units. vinitial =

Homework Answers

Answer #1

(a)

the voltage across the capacitor is

V = E d = 5.0 * 10^5 ( 1.6 * 10^-3) = 800 V

(b)

The capacitence of the capcaitor is

C= eo A/ d

= 8 * 85 * 10^-12 ( pi ( 2.3 * 10^-2 m/2)^2/1.6 * 10^-3

= 2.297 * 10^-12 F

Q= CV = 2.297 * 10^-12 ( 800 V) = 1.838 * 10^-9 C = 1.838 nC

Q = 1.838 nC

(c)

Apply conservation of mechanical energy

delK + del U = 0

del K = - del U

1/2 m vf^2 - 1/2 m vi^2 = -q del V

vf^2 - vi^2 =- 2 q del V/m

vi^2 = vf^2 -2 q del V/m

= ( 2.4 * 10^7)^2 - 2 ( 1.6 * 10^-19) ( 800 V)/9.11 * 10^-31

vi = 1.71 * 10^7 m/s

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