question_answer

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?

Answer:

                                             The original force between the two charges is                                 \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q\times q}{{{r}^{2}}}\] When the individual charges and the distance between them are doubled, the force becomes \[F'=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2q\times 2q}{{{(2r)}^{2}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q\times q}{{{r}^{2}}}=F\] Hence the force will remain same.