Two 2.3-cm-diameter-disks spaced 2.0 mm apart form a parallel-plate capacitor. 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 4.6×105 V/m .

a) What is the voltage across the capacitor?

= 920 V (Correct Answer, don't need)

b) How much charge is on each disk?

It asks to separate your answers by a comma, implying there are two separate answers. I got 1.69x10^-9 C, and that was incorrect.

c) An electron is launched from the negative plate. It strikes the positive plate at a speed of2.0×107 m/s . What was the electron's speed as it left the negative plate?

Any help is appreciated, and if you could show your work please do! Thanks!

Explanation / Answer

(A) V = E*d = (4.6e5 V/m) x (2e-3 m) = 9.2e2 V = 920 V

(B) Gauss' law says that flux across a surface (E x area) is proportional to the charge enclosed by the surface. Choose one disc (no, I don't care which one) and calculate flux. There's only an E-field on one side, more or less, so
E (pi r^2) = Q / e0
where e0 is the permittivity. No dielectric in this capacitor, so e0 is the permittivity of free space, 8.85e-12 F/m:
Q = (8.85e-12 F/m) (4.6e5 V/m)*pi*(1.15e-2 m)^2 = 1.69 nC

(C) The e's kinetic energy at impact was 1/2 m v2^2, where m is the electron's mass and little v2 is its impact velocity. The energy it gained from the field was qV, where big V is the capacitor voltage and q is the charge of an electron. 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 = 4*10^14 - 2*1.6*10^-19*920/9.1*10^-31 = 0.76*10^14

v1 = 8.7*10^6 m/s

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