A solid sphere of radius 0.05 m is uniformly charged with a volume charge density $\rho =1.0\times 10^{-5}{\text{C/m}}^3$. Calculate the magnitude of the electric field at a point 0.08 m from the centre of the sphere (outside the sphere).

Step-by-step solution

1. For a uniformly charged sphere, the total charge inside is $Q=\frac{4}{3}\pi R^3\rho$.
2. Insert $R=0.05\text{m}$ and $\rho =1.0\times 10^{-5}{\text{C/m}}^3$ to find $Q=5.24\times 10^{-9}\text{C}$.
3. The electric field outside a spherical charge distribution is $E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$, where $r=0.08\text{m}$.
4. Compute $\frac{1}{4\pi {\epsilon }_0}=9.0\times 10^9{\text{N⋅m}}^2{\text{/C}}^2$.
5. Evaluate $E=9.0\times 10^9\times \frac{5.24\times 10^{-9}}{(0.08{)}^2}$.
6. Simplify to obtain $E\approx 7.36\times 10^3\text{N/C}$, which rounds to $7.4\times 10^3\text{N/C}$.

Concept tested: Electric field outside a uniformly charged sphere

Published 4 October 2026 · How this solution was prepared and checked