A parallel-plate capacitor consists of two plates, each with an area of 25 cm2 separated by 3.0 mm. The charge on the capacitor is 9.8 nC. A proton is released from rest next to the positive plate. How long does it take for the proton to reach the negative plate?

Electric Field between capacitor plates = Q/(A*E0)

Q = 9.3*10-9 C

A = 25*10-4 m2

E0 = 8.854*10-12 F/m

Electric Field = (9.3*10-9)/(25*10-4*8.854*10-12) = 4.2015*105 N/C

Force on Proton = (1.602 x 10-19)*(4.2015*105) = 6.7224*10-14 N

Acceleration of proton = F/m = (6.7224*10-14)/(1.67262178 10-27) = 4.019*1013 m/s2

Now, v2 = 2*a*s = 2* 4.019*1013*3*10-3

v = 4.91068*105 m/s

Now, v = a*t

t = v/a = (4.91068*105)/(4.019*1013) = 1.2218664*10-8 sec

Time = 1.222*10-8 seconds

answer; equivalent to a one(1) horsepower motor;