Two tiny spheres carrying charges \(1.8 \mu \mathrm{C}\) and \(2.8 \mu \mathrm{C}\) arc located at \(40 \mathrm{~cm}\) apart. The potential at the mid - point of the line joining the two charges is

Charge on first sphere, \(q_1=1.8 \mu \mathrm{C}=1.8 \times 10^{-6} \mathrm{C}\)
Charge on second sphere, \(q_2=2.8 \mu \mathrm{C}=2.8 \times 10^{-6} \mathrm{C}\)
Distance between the two spheres, \(r=40 \mathrm{~cm}\)
Distance between mid-point and each sphere \(r_1=r_2=20 \mathrm{~cm}=0.2 \mathrm{~m}\)
\(\therefore\) Electric potential at mid-point due to both charged sphere
\(\begin{aligned}V & =V_1+V_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_1}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_2} \\& =9 \times 10^9 \times \frac{1.8 \times 10^{-6}}{0.2}+9 \times 10^9 \times \frac{2.8 \times 10^{-6}}{0.2} \\& =9 \times 10^3(9+14)=2.07 \times 10^5 \mathrm{~V} \approx 2.1 \times 10^5 \mathrm{~V}\end{aligned}\)