Two spheres of electric charges \( +2 \mathrm{nC} \) and \( -8 \mathrm{nC} \) are placed at a distance ' \( \mathrm{d} \) ' apart. If they are allowed to touch each other, what is the new distance between them to get a repulsive force of same magnitude as before?

Given, two charges \( +2 \mathrm{nC} \) and \( -8 \mathrm{nC} \) are placed distance \( \mathrm{d} \) apart. Therefore,
\(|F|=\frac{1}{4 \Pi \varepsilon_{0}} \frac{\left(2 \times 10^{-9}\right)\left(8 \times 10^{-9}\right)}{d^{2}}=\frac{k\left(16 \times 10^{-18}\right)}{d^{2}}\)
When the charges are joined together, then
\(\left|F^{\prime}\right|=\frac{1}{4 \Pi \varepsilon_{0}} \frac{9 \times 10^{-18}}{d^{12}}=\frac{k\left(9 \times 10^{-18}\right)}{d^{12}}\)
Now,
\(|F|=\left|F^{\prime}\right| \Rightarrow \frac{k\left(16 \times 10^{-18}\right)}{d^{2}}=\frac{k\left(9 \times 10^{-18}\right)}{d^{12}} \)
\( \Rightarrow \frac{16}{d^{2}}=\frac{9}{d^{12}} \)
\(\Rightarrow d^{12}=\frac{9}{16} d^{2} \)
\(\Rightarrow d^{1}=\frac{3}{4} d\)
Therefore, in order to have same repulsive force as before the new distance between them should be \( \frac{3}{4} d \)