The electric field at a distance of 0.20 m from a point charge q is \( 1.35 \times 10^{3}\,\mathrm{N\,C^{-1}} \). The electric potential at the same distance will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2.7 \times 10^{+2}\,\mathrm{V} \)
For a point charge the field and the potential at distance \( r \) are
\( \displaystyle E = \frac{kq}{r^{2}} \)
\( \displaystyle V = \frac{kq}{r} \)
Dividing the potential by the field gives \( V = E\,r \).
\( \displaystyle V = 1.35\times10^{3}\times0.20 = 270\,\mathrm{V} = 2.7\times10^{2}\,\mathrm{V} \)