The electric potential at a point P due to a point charge of 9 μC located 10 cm away from point P is

$8.1 × 10^5 V$

The correct answer is Option (4) → $8.1 × 10^5 V$

$\text{Given: } q = 9~\mu\text{C} = 9 \times 10^{-6}~\text{C},~ r = 10~\text{cm} = 0.1~\text{m}$

$\text{Electric potential due to a point charge: } V = \frac{k q}{r}$

$k = 9 \times 10^9~\text{Nm}^2/\text{C}^2$

$V = \frac{9 \times 10^9 \cdot 9 \times 10^{-6}}{0.1}$

$V = \frac{81 \times 10^3}{0.1} = 8.1 \times 10^5~\text{V}$

$\text{Answer: } V = 8.1 \times 10^5~\text{V}$