Question

If the distance between two equal point charges is doubled and their individual charges are also doubled, what would happen to the force between them ?

• A.
  Becomes double
• B.
  Becomes half
• C.
  Remains same
• D.
  Becomes four times

Answer 1

The correct option is C

Remains same
$F=\frac{1}{4\pi {ϵ}_0}\frac{q\times q}{r^2}\text{again,}{F}^{\prime }=\frac{1}{4\pi ϵ}\frac{\left(2q\right)\left(2q\right)}{(2r{)}^2}\text{or}{F}^{\prime }=\frac{1}{4\pi {ϵ}_0}\frac{4q^2}{4r^2}=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{r^2}=F$

So, the force will remain the sme.