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?
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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