Two identical conducting spheres P and S with charge Q on each, repel each other with a force $16 \mathrm{~N}$. A third identical uncharged conducting sphere $\mathrm{R}$ is successively brought in contact with the two spheres. The new force of repulsion between $\mathrm{P}$ and $\mathrm{S}$ is :
Solution
<p>$$\begin{aligned}
& F_1=\frac{K Q^2}{r^2}=16 \mathrm{~N} \\
& F_2=\frac{K\left(\frac{Q}{2}\right)\left(\frac{3}{4}\right)}{r^2}=\frac{3}{8} \times 16=6 \mathrm{~N}
\end{aligned}$$</p>
<p>Final charges on spheres are $\frac{Q}{2}$ and $\frac{3 Q}{4}$.</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
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