Question

The electric field strength is 5.50×10⁴ N/C inside a parallel-plate capacitor with a 2.00 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.

Answer 1

Using Work-energy theorem:

W = dKE

W = Work-done by electric field = q*dV

dV = Potential difference between plates = E*d

E = Electric field between plates = 5.50*10^4 N/C

d = plate separation = 2.00 mm = 2.00*10^-3 m

dKE = KEf - KEi

KEi = Initial KE of proton = 0, since proton is released from rest

KEf = (1/2)*m*Vf^2

So,

W = dKE

q*E*d = (1/2)*m*Vf^2 - 0

Vf = sqrt (2*q*E*d/m)

Using given values:

Vf = sqrt (2*1.6*10^-19*5.50*10^4*2.00*10^-3/(1.67*10^-27))

Vf = 145182.1 m/s

Vf = final speed of proton = 1.45*10^5 m/s

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