Two point charges are placed on the x‑axis. A charge $+3\mu \text{C}$ is at the origin and a charge $-2\mu \text{C}$ is at $x=0.20\text{m}$. Calculate the electric potential at the point $P$ located at $x=0.50\text{m}$ on the same axis. 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 expression for potential due to a point charge: $V=k\frac{q}{r}$.
2. Find the distance from each charge to point $P$: $r_1=0.50\text{m}$ from the $+3\mu \text{C}$ charge, and $r_2=0.30\text{m}$ from the $-2\mu \text{C}$ charge.
3. Calculate the contribution of each charge: $V_1=k\frac{+3\times 10^{-6}}{0.50}=9\times 10^9\times 6\times 10^{-6}=5.4\times 10^4\text{V}$, $V_2=k\frac{-2\times 10^{-6}}{0.30}=9\times 10^9\times (-6.67\times 10^{-6})=-6.0\times 10^4\text{V}$.
4. Add the contributions (since potential is a scalar): $V=V_1+V_2=5.4\times 10^4-6.0\times 10^4=-6.0\times 10^3\text{V}$.
5. Convert to proper units: $-6.0\times 10^3\text{V}=-6.0\text{kV}$. (If sign convention is required, the magnitude is $6.0\text{kV}$.)

Concept tested: Potential of point charges

Published 6 September 2026 · How this solution was prepared and checked