A charged particle (q = −8.0 mC), which moves in a region where the only force acting on the particle is an electric force, is released from rest at point A. At point B the kinetic energy of the particle is equal to 4.8 J. What is the electric potential difference VB − VA?

Question

Quizplus · Accepted Answer

We can start by using the conservation of energy principle. The electric force is conservative, so the work done by the force between A and B is equal to the decrease in potential energy:

ΔU = UA - UB = qVB - qVA

Since the particle is released from rest at A, its initial kinetic energy is zero. Therefore, the work done by the electric force is equal to the final kinetic energy:

W = ΔK = K - 0 = 4.8 J

Combining these equations, we get:

qVB - qVA = W
-8.0×10⁻³ VB + 8.0×10⁻³ VA = -4.8

We can solve for VB - VA:

VB - VA = (-4.8/8.0×10⁻³) × (-1) = 6.0×10² V

Therefore, the electric potential difference VB - VA is +0.60 kV (choice B).

A Charged Particle (Q = −8