Answer:
The magnitude of the electric potential decreases.
Explanation:
The electric potential of a positive charge in an electric field is given the equation:
[tex]\displaystyle V = k\left(\dfrac{q}{r}\right)[/tex]
where:
- [tex]k[/tex] = Coulomb's constant (approx. 9.0 × 10⁹)
- [tex]q[/tex] = charge (of the center of the electric field)
- [tex]r[/tex] = distance from the charge
We are looking at the case when the positive charge is moving against the electric field—in other words, when r decreases.
We can see that as [tex]r\downarrow[/tex] ... [tex]V \uparrow[/tex]
because [tex]r[/tex] is in the denominator of a fraction.
This means that the magnitude (or absolute value) of the electric potential decreases.
Note that for an electric field created by a negative charge, this means that the electric potential increases, because it becomes less negative.
