2. The potential difference between the plates of a parallel plate capacitor is

Question

2. The potential difference between the plates of a parallel plate capacitor is 200 V. The circular plates are 5 cm in diameter and are 2mm apart. a) Calculate the electric field between the plates. b) Calculate the amount of charge stored on each plate. o) An electron is released from the negative plate with a speed of 1.2x107 m/s and accelerates towards the positive plate. What is the electron's speed when it strikes the positive plate? d) During this process, what is the electron's change in kinetic energy in electron volt?

Explanation / Answer

(A) E = V/d = 200 / (2 x 10^-3)

= 1 x 10^5 V/m

(B) C = e0 A / d = (8.854 x 10^-12)(pi 0.025^2) / (2 x 10^-3)

C = 8.692 x 10^-12 F

Q = C V = 1.74 x 10^-9 C Or 1.74 nC

(C) delta(KE) = - q deltaV

(9.109 x 10^-31)(v^2 - (1.2 x 10^7)^2)/2 = (1.6 x 10^-19)(200)

v = 1.46 x 10^7 m/s

(D) change in KE = 200 eV

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