A point charge of magnitude $+5\mu \text{C}$ is placed at the origin. Calculate the electric potential at a point $P$ located $15\text{cm}$ away 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. Write the formula for potential due to a point charge: $V=k\frac{q}{r}$.
2. Substitute $q=+5\times 10^{-6}\text{C}$ and $r=0.15\text{m}$ into the formula.
3. Calculate: $V=9\times 10^9\times \frac{5\times 10^{-6}}{0.15}$.
4. Perform the division $\frac{5\times 10^{-6}}{0.15}=3.33\times 10^{-5}$.
5. Multiply by $9\times 10^9$ to obtain $V=3.0\times 10^5\text{V}$.
6. State the result: the potential at $P$ is $3.0\times 10^5\text{V}$.

Concept tested: Potential of a point charge

Published 9 September 2026 · How this solution was prepared and checked