Two equally charged identical metal spheres A and B repel each other with a force $2\times {10}^{-5}\mathrm{N}.$ Another identical uncharged sphere C is touched to A and then placed at the midpoint between A and B. What is the net electric force, in $\mathrm{\mu }$N on C?

Let q be the initial charge on each of spheres A and B, then given

${\mathrm{F}}_{\mathrm{AB}}=2.0\times {10}^{-5}\mathrm{N}$

So from coulomb’s Law

$\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)\frac{{\mathrm{q}}^2}{{\mathrm{r}}^2}=2.0\times {10}^{-5} \dots \left(1\right)$

When the uncharged sphere C is touched with A, the charge on A and C is distributed equally i.e,

${\mathrm{q}}_{\mathrm{A}}={\mathrm{q}}_{\mathrm{C}}=\frac{0+\mathrm{q}}{2}=\frac{\mathrm{q}}{2}$

The arrangement is shown, where C is mid-point of A and B

$\Rightarrow \mathrm{AC}=\mathrm{BC}=\frac{\mathrm{AB}}{2}=\frac{\mathrm{r}}{2}$

The force on C due to A

$\left|{\overrightarrow{\mathrm{F}}}_{\mathrm{CA}}\right|=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_{\mathrm{B}}{\mathrm{q}}_{\mathrm{C}}}{{\mathrm{r}}_{\mathrm{BC}}^2}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left({\displaystyle \frac{\mathrm{q}}{2}}\right)\left({\displaystyle \frac{\mathrm{q}}{2}}\right)}{{\left({\displaystyle \frac{\mathrm{r}}{2}}\right)}^2}$

$\Rightarrow {\mathrm{F}}_{\mathrm{CA}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{{\mathrm{r}}^2}\mathrm{from} \mathrm{A}\to \mathrm{C}$

The force on C due to B is

$\left|{\overrightarrow{\mathrm{F}}}_{\mathrm{CB}}\right|=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_{\mathrm{A}}{\mathrm{q}}_{\mathrm{C}}}{{\mathrm{r}}_{\mathrm{BC}}^2}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\displaystyle \frac{\mathrm{q}}{2}\left(\mathrm{q}\right)}}{{\left({\displaystyle \frac{\mathrm{r}}{2}}\right)}^2}$

$\Rightarrow {\mathrm{F}}_{\mathrm{CB}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2{\mathrm{q}}^2}{{\mathrm{r}}^2}\mathrm{from} \mathrm{B}\to \mathrm{C}$

By principle of superposition, net force on C is

${\overrightarrow{\mathrm{F}}}_{\mathrm{C}}={\overrightarrow{\mathrm{F}}}_{\mathrm{CA}}+{\overrightarrow{\mathrm{F}}}_{\mathrm{CB}}$

$\Rightarrow {\mathrm{F}}_{\mathrm{C}}=\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2{\mathrm{q}}^2}{{\mathrm{r}}^2}-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{{\mathrm{r}}^2}\right) \mathrm{from} \mathrm{B} \mathrm{to} \mathrm{C}$

$\Rightarrow {\mathrm{F}}_{\mathrm{C}}=\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)\frac{{\mathrm{q}}^2}{{\mathrm{r}}^2} \mathrm{from} \mathrm{B} \mathrm{to} \mathrm{C}$

Using (1), we get

$\left|{\overrightarrow{\mathrm{F}}}_{\mathrm{C}}\right|=2.0\times {10}^{-5} \mathrm{from} \mathrm{B} \mathrm{to} \mathrm{C}$

$\Rightarrow \mathrm{F}=20 \mathrm{\mu N}$