Answer:
its potential energy decreases and its electric potential decreases.
Explanation:
Let's consider a radial field for simplicity. We have:
- The electric potential of the field is given by:
[tex]V=\frac{kQ}{r}[/tex]
where
k is the Coulomb's constant
Q is the charge source of the field
r is the distance from Q
We see that the electric potential decreases as we move away from the source. If we consider a positive charge q moving in the direction of the electric field, this charge q will move away from the charge Q (because the field lines generated by the positive particle Q point away from the particle), so the electric potential will decrease.
- The potential energy of the moving charge is given by
[tex]U=qV[/tex]
where q is the magnitude of the charge. As we said previously, V is decreasing while the charge is moving in the direction of the field, so since U is directly proportional to V, U will decrease as well.
