The electric field strength is 520104 NC inside a parallelpl

The electric field strength is 5.20×104 N/C inside a parallel-plate capacitor with a 1.80 mm spacing. A proton is released from rest at the positive plate.

What is the proton\'s speed when it reaches the negative plate?

Solution

Potential diff between plates, V = E.d

V = (5.20x 10^4) (1.80 x 10^-3) = 93.6 Volt

and Gain in KE = q deltaV   { work done by force on charge}

m v^2 /2 - 0   = (1.6 x 10^-19) (93.6)

(1.67 x 10^-27) v^2 /2 = 149.76 x 10^-19

v = 133922.85 m/s