A solid sphere of radius 0.10 m carries a uniform volume charge density $\rho =2.0\times 10^{-5}{\text{C/m}}^3$. Find the magnitude of the electric field at a point 0.08 m from the center (inside the sphere).

Step-by-step solution

1. Choose a spherical Gaussian surface of radius $r=0.08\text{m}$ concentric with the sphere.
2. Gauss's law gives $\Phi =E,4\pi r^2=Q_{\text{enc}}/{\epsilon }_0$.
3. The enclosed charge is $Q_{\text{enc}}=\rho ,\frac{4}{3}\pi r^3$.
4. Equate $E,4\pi r^2=\frac{\rho ,\frac{4}{3}\pi r^3}{{\epsilon }_0}$ and solve for $E$ to get $E=\frac{\rho r}{3{\epsilon }_0}$.
5. Substitute the numbers: $\rho =2.0\times 10^{-5}{\text{C/m}}^3$, $r=0.08\text{m}$, ${\epsilon }_0=8.854\times 10^{-12}{\text{C}}^2/({\text{N⋅m}}^2)$.
6. Compute $\rho r=1.6\times 10^{-6}{\text{C/m}}^2$ and $3{\epsilon }_0=2.6562\times 10^{-11}{\text{C}}^2/({\text{N⋅m}}^2)$.
7. Divide to obtain $E=1.6\times 10^{-6}/2.6562\times 10^{-11}\approx 6.0\times 10^4\text{N/C}$, radially outward.

Concept tested: Gauss's law for symmetric charge distributions

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