A proton moves in a constant electric field from point A to point B. The magnitude of the electric field is 6.4 × 10⁴ N/C; and it is directed as shown in the drawing, the direction opposite to the motion of the proton. If the distance from point A to point B is 0.50 m, what is the change in the proton's electric potential energy, EPE_A - EPE_B?

Question

Quizplus · Accepted Answer

The change in electric potential energy of a charged particle moving in an electric field is given by: 
ΔU = qΔV, where q is the charge of the particle and ΔV is the change in electric potential. 
The electric potential difference, ΔV, is given by: 
ΔV = V_B - V_A = -Ed, where E is the magnitude of the electric field, d is the distance moved by the particle, and the negative sign indicates that the electric field is opposite to the direction of motion of the particle. 
Thus, ΔU = qΔV = -qEd. 
In this problem, the charge of the proton is q = +1.6 × 10^-19 C, the magnitude of the electric field is E = 6.4 × 10^4 N/C, and the distance traveled by the proton is d = 0.5 m. 
Therefore, ΔU = -qEd = -(1.6 × 10^-19 C)(6.4 × 10^4 N/C)(0.5 m) = -5.12 × 10^-15 J. 
The answer is D, -5.1 × 10^(-15) J.

A Proton Moves in a Constant Electric Field from Point