1. If the distance between two negative point charges is increased by a factor of three, the resultant potential energy is what factor times the initial potential energy?
ans: 1/3; PE = k*q1*q2/d.
2. A proton (+1.6 x10-19 C) moves 10 cm on a path parallel to the direction of a uniform electric field of strength 3.0 N/C. How much work is done on the proton by the electrical field?
ans: 4.8 x10-20 J; W = f*d = q*E*d.
3. If an electron is accelerated from rest through a potential difference of 1200 V, find its approximate velocity at the end (qe = -1.6 x10-19 C; me = 9.1 x10-31 kg)
ans: 2.1 x107 m/s; KE = -PE = q*V = m*v*v/2.