A uniform electric field,with a magnitude of 500 N/C,is directed parallel to the positive x-axis.If the potential at x = 3.0 m is 1,800 V,what is the change in potential energy of a proton as it moves from x = 4 m to x = 3 m? (q_p = 1.6 × 10−¹⁹ C)

Question

Quizplus · Accepted Answer

The electric potential difference between two points in an electric field is given by the formula ΔV = -Ed, where E is the electric field strength and d is the distance between the two points. In this case, the proton is moving from x = 4 m to x = 3 m, which means it is moving opposite to the direction of the electric field. Therefore, the change in potential energy of the proton is given by ΔPE = qΔV, where q is the charge of the proton. Substituting in the given values, we get:

ΔPE = (1.6 × 10^-19 C)(-(500 N/C)(1 m)) = -8 × 10^-17 J

Notice that the distance moved is -1 m, since the proton is moving opposite to the electric field direction, and the electric potential difference is negative because the proton is moving from a point of higher potential (x = 4 m) to a point of lower potential (x = 3 m). Therefore, the proton loses potential energy as it moves. The answer is B, 2.88 × 10^-16 J, which is the same as -8 × 10^-17 J rounded to two significant figures.

A Uniform Electric Field,with a Magnitude of 500 N/C,is Directed