Two point charges \((A\) and \(B)+4 q\) and \(-4 q\) are placed along a line separated by a distance ' \(r\) '. Force acting between them is F. If \(25 \%\) of charge from point A is transferred to that at
point \(B\), the force between the charges now becomes

Force acting between given charges \(+4 q\) and \(-4 q\) is, \(F=\frac{-16 q^2}{4 \pi \varepsilon_0 r^2}\)...(i)
When \(25 \%\) of charges ảre transferred, charge on point A becomes,
\(q_1=+4 q-0.25(+4 q)=+3 q\)
Charge on point B becomes,
\(q_2=-4 q+0.25(+4 q)=-3 q\)
\(\therefore \quad\) The new force between points \(A\) and \(B\) will be,
\(\mathrm{F}^{\prime}=\frac{(3 \mathrm{q}) \times(-3 \mathrm{q})}{4 \pi \varepsilon_0 \mathrm{r}^2}=\frac{-9 \mathrm{q}^2}{4 \pi \varepsilon_0 \mathrm{r}^2}\)
Multiplying and dividing by 16 , \(F^{\prime}=\frac{9}{16} \times\left(\frac{-16 q^2}{4 \pi \varepsilon_0 r^2}\right)=\frac{9}{16} F \ldots[\operatorname{From}(i)]\)