Answer to Question #309267 in Physics for Mave

Question

A charge of q = -5.3 nC is accelerated from rest through a potential

difference of 25 V. What is the change in electric potential energy of

the charge?

Answer 1

The change in electric potential energy of the charge

$$\Delta E_p=qV\\ =-5.3*10^{-9}*25=-1.3*10^{-7}\:\rm J$$