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 V_B − V_A? A) −0.60 kV B) +0.60 kV C) +0.80 kV D) −0.80 kV E) +0.48 kV

Question

Quizplus · Accepted Answer

Using conservation of energy, we have:
qVS1U1B0 - qVS1U1B1 - K = 0
Where K is the kinetic energy of the charged particle at point B.
Solving for VS1U1B1, we have:
qVS1U1B1 = qVS1U1B0 - K
VS1U1B1 = VS1U1B0 - K/q
Note that q is negative, so we have:
VS1U1B1 - VS1U1B0 = - K/|q|
Substituting K = 4.8 J and q = -8.0x10^-6 C, we get:
VS1U1B1 - VS1U1B0 = - 4.8 J/|(-8.0x10^-6 C)| = +0.60 kV
Therefore, the answer is B: +0.60 kV.

A Charged Particle (Q = −8