Two equally charged metal spheres $A$ and $B$ repel each other with a force of $4 imes 10^{-5} \mathrm{~N}$. Another identical uncharged sphere $C$ is touched to $A$ and then placed at the mid- point of the line joining the spheres $A$ and $B$. The net electric force on the sphere $C$ is

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Two equally charged metal spheres $A$ and $B$ repel each other with a force of $4 	imes 10^{-5} \mathrm{~N}$. Another identical uncharged sphere $C$ is touched to $A$ and then placed at the mid- point of the line joining the spheres $A$ and $B$. The net electric force on the sphere $C$ is

Aicharya · Accepted Answer

Force, $\quad F=\frac{k q^2}{d^2}=4 	imes 10^{-5} \mathrm{~N}$ Now, $A$ is touched by $C$, then; Charge on $C=q / 2$ Charge on $A=q / 2$ So, force on $C=\mathbf{F}_A+\mathbf{F}_B$ $ \begin{aligned} & =\frac{k q / 2 \cdot q / 2}{(d / 2)^2} \hat{\mathbf{r}}_{A C}+\frac{k q \cdot q / 2}{(d / 2)^2} \hat{\mathbf{r}}_{B C} \ & \left.=\frac{k q^2 / 4}{d^2 / 4}-\frac{k q^2 / 2}{d^2 / 4} \quad \quad 	ext { as } \hat{\mathbf{r}}_{A C}=-\hat{\mathbf{r}}_{B C}ight] \ & =\frac{k q^2}{d^2}(1-2)=-\frac{k q^2}{d^2}=-4 	imes 10^{-5} \mathrm{~N} \end{aligned} $ So, force is of same magnitude but in opposite direction, i.e from $C$ to $A$ as suggested by minus sign

Two equally charged metal spheres $A$ and $B$ repel each other with a force of $4 \times 10^{-5}…

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