The electric potential at the centre of a square of side $\sqrt{2} m$ having charges 90 μC, -60 μC, 30 μC and -40 μC at the four corners of the square is

$1.8 × 10^5 V$

The correct answer is Option (1) → $1.8 × 10^5 V$

Electric potential at the centre of a square due to a point charge:

$V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}$

Distance from center to a corner of a square of side $a$: $r = \frac{a}{\sqrt{2}}$

Given: $a = \sqrt{2} \, \text{m} \Rightarrow r = \frac{\sqrt{2}}{\sqrt{2}} = 1 \, \text{m}$

Charges at corners: $q_1 = 90 \, \mu\text{C}$, $q_2 = -60 \, \mu\text{C}$, $q_3 = 30 \, \mu\text{C}$, $q_4 = -40 \, \mu\text{C}$

Net potential at center (scalar sum):

$V = \frac{1}{4 \pi \epsilon_0} \frac{q_1 + q_2 + q_3 + q_4}{r}$

$q_\text{total} = 90 - 60 + 30 - 40 = 20 \, \mu\text{C} = 20 \times 10^{-6} \, \text{C}$

$r = 1 \, \text{m} \Rightarrow V = \frac{1}{4 \pi \epsilon_0} \cdot 20 \times 10^{-6}$

$\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2$

$V = 9 \times 10^9 \cdot 20 \times 10^{-6} = 1.8 \times 10^5 \, \text{V}$

V $\approx 1.8 \times 10^5 \, \text{V}$