A point charge of +5 µC is placed at the origin. Find the magnitude of the electric field at the point (0.12 m, 0, 0) on the x–axis.

Step-by-step solution

1. Use the formula for the electric field of a point charge: $E=\frac{k|q|}{r^2}$, where $k=8.99\times 10^{9}N\cdot m^2/C^2$, $q=5\times 10^{-6}C$, and $r=0.12m$.
2. Calculate $r^2=0.12^2=0.0144m^2$.
3. Compute $E=\frac{8.99\times 10^9\times 5\times 10^{-6}}{0.0144}=1.48\times 10^{5}N/C$.
4. The field points radially away from the positive charge, i.e., in the +x direction.

Concept tested: Electric field of a point charge

Published 13 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked