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
Answer & explanation
Correct answer: option 2
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:
$V = 1.35 \times 10^3 \cdot 0.20$
$V = 270 \, \text{V}$
${V \approx 2.7 \times 10^2 \, \text{V}}$