A particle (m = 2.0 μg, q = −5.0 μC) has a speed of 30 m/s at point A and moves (with only electric forces acting on it) to point B where its speed is 80 m/s. Determine the electric potential difference VA − VB.

Question

Quizplus · Accepted Answer

We can use the conservation of energy principle since there is only electric force acting on the particle.

The change in kinetic energy of the particle is

ΔK = Kf - Ki = (1/2)mvf^2 - (1/2)mvi^2

Substituting given values, we get

ΔK = (1/2)(2.0×10^-9 kg)(80 m/s)^2 - (1/2)(2.0×10^-9 kg)(30 m/s)^2
    = 5.8×10^-16 J

This change in kinetic energy is equal in magnitude to the work done by the electric force during the particle's motion. Thus,

W = ΔK = qΔV
where ΔV = VA - VB is the potential difference between points A and B.

Solving for ΔV, we get

ΔV = ΔK/q = (5.8×10^-16 J)/(-5.0×10^-6 C)
     = -1.16×10^2 V
     = -1.1 kV (to two significant figures)

Therefore, the correct answer is C.

A Particle (M = 2