Question

He electric potential at a certain distance from a point charge can be represented by v. what is the value of the electric potential at twice the distance from the point charge?

Answer 1

Coulomb's Law for the electric field due to a point charge is
$E=k_{e} \frac{q}{r^{2}}$
where
E = electric field, V
$k_{e}$= Coulom's constant, 8.98755 x 10²
q = point charge, C
r =  distance from the pont charge, m

At distance, r, let the electric potential be
$v=k_{e} \frac{q}{r^{2}}$

At twice the distance from q, the electric potential is
$v'=k_{e} \frac{q}{(2r)^{2}} \\ = \frac{1}{4} k_{e} \frac{q}{r^{2}} \\ = \frac1x}{4}v$

Answer:
At twice the distance from the charge, the electric decreases by a factor of 4.