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

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Solution

Initial force $\Rightarrow F = K \frac{q q}{r^{2}}$
Final force $\Rightarrow F = K \frac{(2 q) (2 q)}{(2 r)^{2}} = \frac{4 K q q}{4 r^{2}} = \frac{K q q}{r^{2}}$
Thus, the force will remain unchanged.

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

Initial force ⇒F=Kqqr2Final force ⇒F=K(2q)(2q)(2r)2=4Kqq4r2=Kqqr2Thus, the force will remain unchanged.

Initial force $\Rightarrow F = K \frac{q q}{r^{2}}$
Final force $\Rightarrow F = K \frac{(2 q) (2 q)}{(2 r)^{2}} = \frac{4 K q q}{4 r^{2}} = \frac{K q q}{r^{2}}$
Thus, the force will remain unchanged.