JEE MAIN - Physics (2024 - 6th April Evening Shift - No. 13)
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 :
1 N
6 N
12 N
4 N
Explanation
$$\begin{aligned} & F_1=\frac{K Q^2}{r^2}=16 \mathrm{~N} \\ & F_2=\frac{K\left(\frac{Q}{2}\right)\left(\frac{3}{4}\right)}{r^2}=\frac{3}{8} \times 16=6 \mathrm{~N} \end{aligned}$$
Final charges on spheres are $$\frac{Q}{2}$$ and $$\frac{3 Q}{4}$$.
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