The electric field strength is 550 times 104 m NC inside a

The electric field strength is 5.50 times 10^4 \\({\ m N/C}\\) inside a parallel-plate capacitor with a 1.50 \\({\ m mm}\\) spacing. A proton is released from rest at the positive plate. What is the proton\'s speed when it reaches the negative plate? Express your answer with the appropriate units.

Solution

V = E*d = 5.5*10^4*1.5*10^-3 = 82.5 V

Using energy conservation method:

KE1 + UE1 = KE2 + UE2

0.5*m1*v1^2 + qV1 = 0.5*m1*v2^2 + q*V2

V2 = 0 (potential)

v1 = 0 (velocity)

v2^2 = 2*q*V1/m

v2 = sqrt(2*1.6*10^-19*82.5/(1.67*10^-27)) = 1.26*10^5 m/sec

Let me know if you have any doubt.