A point charge of +5 µC is placed at the origin. Find the magnitude of the electric field at a point that lies on the y‑axis 0.30 m away from the origin.

Step-by-step solution

1. Step 1: Identify the formula for the electric field due to a point charge: $E=\frac{k,|Q|}{r^2}$, where $k=9.0\times 10^9,{\text{N m}}^2{\text{/C}}^2$.
2. Step 2: Substitute the given values: $Q=5\times 10^{-6},\text{C}$ and $r=0.30,\text{m}$ into the formula.
3. Step 3: Compute the numerator: $k|Q|=9.0\times 10^9\times 5\times 10^{-6}=4.5\times 10^4,{\text{N m}}^2\text{/C}$.
4. Step 4: Compute the denominator: $r^2=0.30^2=0.09,{\text{m}}^2$.
5. Step 5: Divide to get the field: $E=\frac{4.5\times 10^4}{0.09}=5.0\times 10^5,\text{N/C}$.

Concept tested: Electric field of point charge

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked