Question

The force between two point charges separated by a certain distance is $F$. If each charge is double and the distance between them is also doubled then the force would be

• A. $2F$
• B. $F$
• C. $F/2$
• D. $4F$

Answer 1

The correct option is B $F$

$\text{Step 1: Initial Force}$

Initially, the force between charges say $q_1$ and $q_2$ seperated with distance $r$ is :

$F=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$
$\text{Step 2: New force calculation}$

When the charge is doubled and the distance between charges is also doubled,

i.e $q_1^{\prime }=2q_1,q_2^{\prime }=2q_2,r^{\prime }=2r$
The new force will be :

$F^{\prime }=\frac{1}{4\pi {ϵ}_0}\frac{q_1^{\prime }{q}_2^{\prime }}{r^{\prime 2}}$

$F^{\prime }=\frac{1}{4\pi {ϵ}_0}\frac{\left(2q_1\right)\left(2q_2\right)}{(2r{)}^2}=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}=F$
Hence, the force will remain the same.

Hence, Option(B) is correct.