A solid sphere of radius 0.05 m carries a uniform volume charge density ρ=3.0×10⁻⁶ C/m³. Find the magnitude of the electric field at a point 0.08 m from the center (outside the sphere).

Step-by-step solution

1. For $r>R$, the field is as if all the charge were concentrated at the centre: $E=\frac{1}{4pivarepsilo{n}_0}\frac{Q_{\text{enc}}}{r^2}$.
2. Calculate the total charge: $Q_{\text{enc}}=\rho \cdot \frac{4}{3}\pi R^3=3.0\times 10^{-6}\cdot \frac{4}{3}\pi (0.05{)}^3=1.57\times 10^{-7}C$.
3. Insert into the formula: $E=\frac{9.0\times 10^9\times 1.57\times 10^{-7}}{(0.08{)}^2}=4.12\times 10^{3}N/C$.
4. Direction is radially outward from the centre.
5. Hence the field magnitude is $4.12\times 10^{3}N/C$.

Concept tested: Gauss's law – field outside uniformly charged sphere

Published 13 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked