Question

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 5.50×104 m/s .

What will be the proton's final speed if the experiment is repeated with double the amount of charge on each capacitor plate?

Answer 1

using conservation of energy

qV = 0.5 m v^2... (i)

volatge across capacitor

V = Q/C

putting in (i)

q (Q/C) = 0.5 m v^2

from the above equation we can conclude that

Q is directly proportional to velocity

so

(v2 /v1) ^2 = Q2/ Q1

v2 / v1 = sqrt(2)

v2 = v1 sqrt (2)

v2 = 5.5* 10^4* sqrt (2)

v2 = 7.778* 10^4 m/s

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