A point charge of $+1\mu \text{C}$ is placed at the origin. Calculate the magnitude of the electric field at the point $P$ located $0.20\text{m}$ on the positive $z$‑axis.

Step-by-step solution

1. Write Coulomb's law for the field of a point charge: $E=k\frac{|q|}{r^2}$, where $k=9\times 10^9{\text{Nm}}^2/{\text{C}}^2$.
2. Substitute $q=1\times 10^{-6}\text{C}$ (convert $\mu \text{C}$ to C) and $r=0.20\text{m}$ into the formula.
3. Calculate $r^2=(0.20{)}^2=0.040{\text{m}}^2$.
4. Compute $E=9\times 10^9\times \frac{1\times 10^{-6}}{0.040}=9\times 10^9\times 2.5\times 10^{-5}=2.25\times 10^5\text{N/C}$.
5. Check the arithmetic: $2.25\times 10^5\text{N/C}=4.5\times 10^4\text{N/C}$ after correcting the factor (the correct division gives $4.5\times 10^4$).
6. State that the direction of the field is along $+z$ because the charge is positive.

Concept tested: Electric field of a point charge

Published 19 September 2026 · How this solution was prepared and checked