Answer is : (B) 2.1 x 105V
Here q1 = 1.8μc = 1.8 x 10-6 C, q2 = 2.8μC = 2.8 x 10-6 C
Distance between the two spheres = 40 cm = 0.4 m
For the mid-point r1 = r2 = 0.40/2 = 0.2 m
Potential at O
V = V1+V2 \(=\frac1{4\pi\varepsilon_0}\left[\frac{q_1}{r_1}+\frac{q_2}{r_2}\right]=\frac{9\times10^9(1.8+2.8)\times10^{-6}}{0.2}\) = 2.1 x 105V