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Q. If charge and distance between two charges are reduced to half. Force between them

Delhi UMET/DPMTDelhi UMET/DPMT 2008Electric Charges and Fields

Solution:

As per Coulomb's law $F = k \frac{q_{1} q_{2}}{r^{2}}$
According to question,
$F'= k \frac{\left(q_{1} / 2\right)\left(q_{2} / 2\right)}{(r / 2)^{2}}$
$=k \frac{q_{1} q_{2}}{r^{2}}=F$
Thus, force between them will remain same.