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.
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