Question

Two positive point charges Q and 2Q are separated by a distance R. If the charge Q experiences a force of magnitude F when the separation is R, what is the magnitude of the force on the charge 2Q when the separation is 2R

Answer 1

Answer:

$F'=\frac{1}{4}F$

Explanation:

According to Coulomb's law, the magnitude of each of the electric forces experiences by the two point charges is directly proportional to the product of the magnitude of both charges ($Q,2Q$) and inversely proportional to the square of the distance(R) that separates them:

$F=\frac{kQ(2Q)}{R^2}$

We have $R'=2R$

$F'=\frac{kQ(2Q)}{R'^2}\\F'=\frac{kQ(2Q)}{(2R)^2}\\F'=\frac{1}{4}\frac{kQ(2Q)}{R^2}\\F'=\frac{1}{4}F$