A point charge of magnitude $+2\mu \text{C}$ is placed at the origin. Calculate the electric potential at a point $P$ located $20\text{cm}$ from the charge. Take the potential to be zero at infinity. Use $k=9\times 10^9{\text{N m}}^2{\text{C}}^{-2}$.

Step-by-step solution

1. Convert the distance to metres: $r=20\text{cm}=0.20\text{m}$.
2. Write the formula for the potential due to a point charge: $V=\frac{kq}{r}$.
3. Substitute the given values: $V=\frac{9\times 10^9\times 2\times 10^{-6}}{0.20}$.
4. Calculate the numerator: $9\times 10^9\times 2\times 10^{-6}=1.8\times 10^4$.
5. Divide by $0.20$: $V=\frac{1.8\times 10^4}{0.20}=9.0\times 10^2\text{V}$.
6. Thus the potential at point $P$ is $900\text{V}$.

Concept tested: Potential of a point charge

Published 22 September 2026 · How this solution was prepared and checked