Two 2.3-cm-diameter-disks spaced 2.0 mm apart form a parallel-plate capacitor. T
ID: 1656894 • Letter: T
Question
Two 2.3-cm-diameter-disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.1×105 V/m .
What is the voltage across the capacitor?
Express your answer with the appropriate units.
How much charge is on each disk?
Enter your answers numerically separated by a comma. Express your answers in coulombs.
An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.5×107 m/s . What was the electron's speed as it left the negative plate?
Express your answer with the appropriate units.
Explanation / Answer
(A) Voltage
(4.1*10^5) * (2*10^-3 ) = 8.2*10^2 V = 820 V
(B)
From Gauss' law flux across a surface is proportional to the charge enclosed by the surface
E (pi r^2) = Q / e0
Q = (9*10^-12 F/m) (4.1*10^5 V/m) pi (1.15*10^-2 m)^2
Q = 1.532*10^-9 c
C) The e's kinetic energy at impact was 1/2 m v2^2, The energy it gained from the field was qV.
The kinetic energy of the electron when it departed was then
1/2 m v1^2 = 1/2 m v2^2 - qV
v1^2 = v2^2 - 2qV / m
v1^2 = (2.5*10^7)^2 - 2*1.6*10^-19*820/(9.1*10^-31)
v1^2 = 3.3664*10^14
v1 = 1.83*10^7 m/s
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