Question

Two metal spheres each have a radius of 10.0 cm. The centers of the two spheres are 54.1 cm apart. The spheres are initially neutral, but a charge Q is transferred from one sphere to the other, creating a potential difference between the spheres of 107 V. A proton is released from rest at the surface of the positively charged sphere and travels to the negatively charged sphere.

(a) What is the proton's kinetic energy just as it strikes the negatively charged sphere?
( ) eV

(b) At what speed does it strike the sphere?

( ) m/s

Answer 1

Part A.

Using energy conservation between initial and final position of proton,

KEi + PEi = KEf + PEf

KEi = 0, since proton is released from rest

KEf = final KE when proton reaches at negatively charged sphere's location

PEf - PEi = W = q*dV

dV = Potential difference between both spheres = 107 V

q = charge on proton = 1.6*10^-19 C = +e

So,

KEf = KEi + PEi - PEf

KEf = 0 + q*dV

KEf = e*(107 V)

KEf = 107 eV

Part B.

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

Vf = sqrt (2*KEf/m)

KEf = 107 eV = 107*1.6*10^-19 J

So,

Vf = sqrt (2*107*1.6*10^-19/(1.67*10^-27))

Vf = 1.43*10^5 m/s = Speed of proton when it strikes the sphere

Let me know if you've any query.