A capacitor is constructed from two circular plates, each with a radius of 1.5 m

Question

A capacitor is constructed from two circular plates, each with a radius of 1.5 m and separated a distance s = 4 mm. These plates are then charged with a charge magnitude of 1.2 C. An electron is then shot between the plates. initially at location, A, halfway between the plates of a capacitor, moving in the +x direction with a speed of vi = 1.6 × 106 m/s. The electron travels along the path shown and collides with the upper plate at location D. The capacitor is made out of circular plates with a radius of 1.5 m and the distance between the plates, s, is 4 mm. Location B is directly beneath location D, as in the diagram.

A) Which plate, top or bottom, is positively charged? Briefly explain.

B) Calculate the electric field within the capacitor. Express your answer as a 3-component vector.

C) Calculate the potential difference, VD VA between points A and D.

D) Calculate the final speed the electron has just before it collides with the top plate.

Explanation / Answer

Here,

radius ,r = 1.5 m

delta , d = 4 mm

charge , q = 1.2 uC

vi = 1.6 *10^6 m/s

a) As the electron is negatively charged and it will be attracted to positive charge

B)

electric field between the capacitor = Q/(epsilon * area)

electric field between the capacitor = 1.2 *10^-6/(8.854 *10^-12 * pi * 1.5^2)

electric field between the capacitor = 1.92 *10^4 N/C

C)

for the potential difference between D and A

potential difference between D and A = s/2 * E

potential difference between D and A = 0.004/2 * 1.92 *10^4

potential difference between D and A = 38.4 V

D)

let the final speed of electron is v

0.5 * 9.11 *10^-31 * ((v)^2 - (1.6 *10^6)^2) = 1.602 *10^-19 * 38.4

solving for v

v = 4.08 *10^6 m/s

the final speed of the electron is 4.08 *10^6 m/s

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