The electric field at a distance of 0.20 m from a point charge q is $1.35×10^3 N C^{-1}$. The electric potential at the same distance will be

$2.7 × 10^{+2}V$

The correct answer is Option (2) → $2.7 × 10^{+2}V$

Given: Electric field at $r = 0.20 \, \text{m}$ is $E = 1.35 \times 10^3 \, \text{N/C}$

Electric potential at a distance $r$ from a point charge is related to electric field by:

$V = E \cdot r$

Substitute the values:

$V = 1.35 \times 10^3 \cdot 0.20$

$V = 270 \, \text{V}$

${V \approx 2.7 \times 10^2 \, \text{V}}$