Two charges, +20 μC and +40 μC are placed at a separation of 4 cm. The electric potential due to the pair at the middle point of the line joining the two charges is

27 MV

The correct answer is Option (4) → 27 MV

Distance of midpoint from each charge = $\frac{4}{2} = 2 \,\text{cm} = 0.02 \,\text{m}$

Potential due to first charge: $V_1 = \frac{k Q_1}{r} = \frac{9 \times 10^9 \times 20 \times 10^{-6}}{0.02}$

$V_1 = 9 \times 10^9 \times \frac{20 \times 10^{-6}}{0.02} = 9 \times 10^9 \times 10^{-3} = 9 \times 10^6 \,\text{V}$

Potential due to second charge: $V_2 = \frac{k Q_2}{r} = \frac{9 \times 10^9 \times 40 \times 10^{-6}}{0.02}$

$V_2 = 9 \times 10^9 \times \frac{40 \times 10^{-6}}{0.02} = 18 \times 10^6 \,\text{V}$

Total potential: $V = V_1 + V_2 = (9 + 18) \times 10^6 = 27 \times 10^6 \,\text{V}$

Answer: $2.7 \times 10^7 \,\text{V}$