Two spheres carrying charges \( +6 \mu \mathrm{C} \) and \( +9 \mu \mathrm{C} \). separated by a distance \( \mathrm{d} \), experiences a
force of repulsion \( F \). When a charge of \( -3 \mu \mathrm{C} \) is given to both the sphere and kept at the same
distance as before, the new force of repulsion is

Given, charge \(q_{1}=+6 \mu C=6 \times 10^{-6} C ;\) charge \(q_{2}=+9 \mu C\)
\(=9 \times 10^{-6} C\) seperation \(=d\)
Force between charges is given as
\(F=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{1} q_{2}}{d^{2}} \propto \frac{q_{1} q_{2}}{d^{2}}=\frac{6 \times 10^{-6} \times 9 \times 10^{-6}}{d^{2}} \rightarrow(1)\)
When a charge of \(+3 \mu C\) is given to both and separation \(=d\), then
\(q_{1}=+6 \mu C-3 \mu C=3 \mu C=3 \times 10^{-6} C\)
\(q_{2}=+9 \mu C-3 \mu C=6 \mu C=6 \times 10^{-6} C\)
The force between charges is given as
\(F^{\prime}=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{1}^{\prime} q_{2}^{\prime}}{d^{2}} \propto \frac{q_{1}^{\prime} q_{2}^{\prime}}{d^{2}}=\frac{3 \times 10^{-6} \times 6 \times 10^{-6}}{d^{2}} \rightarrow(2)\)
Dividing Eq. \((1)\) by Eq. \((2)\), we get
\(\frac{F}{F}=\frac{6 \times 10^{-6} \times 9 \times 10^{-6}}{3 \times 10^{-6} \times 6 \times 10^{-6}}=3\)
\(\Rightarrow F^{\prime}=\frac{F}{3}\) Therefore, the new force of repulsion is \(\frac{F}{3}\)