A particle (charge = 50 $\mu$C)moves in a region where the only force on it is an electric force.As the particle moves 25 cm from point A to point B,its kinetic energy increases by 1.5 mJ.Determine the electric potential difference,V_B - V_A. A)-50 V B)-40 V C)-30 V D)-60 V E)+15 V

Question

Quizplus · Accepted Answer

We can use the formula for work done by electric force: 
W = -ΔPE = ΔKE 
where W is the work done by the electric force, ΔPE is the change in potential energy, and ΔKE is the change in kinetic energy. Since the only force acting on the particle is electric, we can assume that all the work done on the particle comes from the electric force.

We can also use the formula for electric potential: 
V = PE/q 
where V is the electric potential, PE is the electric potential energy, and q is the charge.

Combining these two formulas, we get: 
W = -qΔV = ΔKE 
where ΔV is the change in electric potential.

Solving for ΔV, we get: 
ΔV = -ΔKE/q 
Substituting the given values, we get: 
ΔV = -(1.5 mJ)/(50 μC) 
ΔV = -30 V

Since the particle moves from point A to point B, the electric potential difference between A and B is positive, and is given by: 
ΔV = VB - VA

Therefore: 
VB - VA = 30 V 
VA - VB = -30 V 
We can see that the only answer choice that is consistent with this result is (C) 30 V.

A Particle (Charge = 50 μ\muμ C)moves in a Region Where the Only Force on It

A Particle (Charge = 50 $\mu$ C)moves in a Region Where the Only Force on It