Question

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? Group of answer choices –0.80 kV +0.80 kV –0.48 kV –0.60 kV +0.60 kV

Answer 1

Answer:

The electric potential difference between two given points is $0.6KV$

Explanation:

Here we know that work done by electrostatic force = change in its kinetic energy

   work done= $4.8J$

since Δv =$V_{B}-V_{A}$ = $\frac{-work \, done \, by \, electrostaticforce}{charge}$

Δv =$V_{B} -V_{A}$= $\frac{-4.8}{-8\times 10^{-3}}$ = $0.6KV$