Two tiny spheres carrying charges $1.8 \mu C$ and $2.8 \mu C$ are located at $40 cm$ apart. The potential at the mid-point of the line joining the two charges is

Here, $q_1=1.8 \mu C=1.8\times {10}^{-6} C, q_2=2.8 \mu C=2.8\times {10}^{-6} C$

Distance between the two spheres$=40 cm=0.4 m$

For the mid-point $r_1=r_2=\frac{0.40}{2}=0.2 m$

Potential at $O,$

$V=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{q_1}{r_1}+\frac{q_2}{r_2}\right]=\frac{9\times {10}^9\left(1.8+2.8\right)\times {10}^{-6}}{0.2}=2.1\times {10}^5 V$