Topic: Electric Potential Energy in a Uniform Electric Field

Sample Problem A point charge of 3.0 nC with a mass of 4.0 g is moved from x =1.0 m to x =1.5 m in an electric field of 5.0 N/C with the same direction as the motion of the charge. (a) How much work is done on the charge by the electric force? (b) What is the change in the potential energy of the charge? (c) Assuming that the charge started from rest, what is its speed at x =1.5 m? Given: q 3.0 nC = 3.0X10-9 C 4.0g - 4.0X10-3 kg E = 5.0 N/C d 1.5 m-1.0 m = 0.5 m Solution: a . W= Fd From Eg. (1.4), F = qE. Therefore, qEd (3.0X 10-9 C)(5.0 N/C)(0.5 m) 7. 5x 10-9 N . m 7.5 x 10-9J b . AU= W - -7.5x10-" JC . Using the work-energy theorem, -K- Ko 1 1 Since vo O (assuming that the charge started from rest), 1 7.5X10-9J (4.0X10 kg)v N 1.9x10 m/sMINITASK #1. Solve the problem below. Show proper solution. A point particle of charge 2.5 x 109 C and a mass of 3.65 x 10-7 kg in uniform electric field directed to the right. It is released from rest and moves to the right. It is released from rest and moves to the right. After it has traveled 1.45 m, its speed is 35 m/s. Find, a. work done on the particle\f