Question

Question 4

The potential energy of a + 0.6C charge decreases from 1.8 J to 0.2 J when it is moved from point A to point B. a. Calculate the potential at point A b. Calculate the potential at point B c. What is the change in the electric potential between these two points?

Answer 1

Given that

charge q=0.6 c

work at point A=>Wa=1.8 J

work at point B=>Wb=0.2 J

basing on the concept of electric potential

now we find the potential point A

Va=Wa/q=1.8/0.6=3 v

now we find the potential point B

Vb=Wb/q=0.2/0.6=0.33 v

now we find the other charge in electric potential

we assume the distance b/w the charges r=1 m

(0.2-1.8)=9*10^9*0.6*q2/1

charge q2=2.963*10^-10 c