A point charge of $+8\mu \text{C}$ is placed at the origin. Find the magnitude of the electric field at the point $(0,0,0.40\text{m})$.

Step-by-step solution

1. Write Coulomb's law for the electric field of a point charge: $E=k\frac{|q|}{r^2}$.
2. Identify the charge magnitude: $|q|=8\times 10^{-6}\text{C}$.
3. Calculate the distance from the charge to the point: $r=0.40\text{m}$.
4. Substitute the values and use $k=9\times 10^9{\text{N⋅m}}^2/{\text{C}}^2$: $E=9\times 10^9\times \frac{8\times 10^{-6}}{(0.40{)}^2}$.
5. Compute the denominator: $(0.40{)}^2=0.16$.
6. Perform the division: $E=\frac{9\times 10^9\times 8\times 10^{-6}}{0.16}=4.5\times 10^5\text{N/C}$.

Concept tested: Electric field of a point charge

Published 10 September 2026 · How this solution was prepared and checked