Answer:
At twice the distance, the electric potential is V/2.
Explanation:
The electric potential at a certain distance d from a point charge q can be represented by V:
V = kq/d .....1
Where q = charge ,d = distance between them, k = coulomb's constant.
When the distance is doubled d1 = 2d
Let V1 represent the electric potential after the distance have been doubled.
V1 = kq/2d = (kq/d)/2
V1 = V/2
Therefore, at twice the distance the electric potential is halved V1 = V/2