A solid insulating cylinder of radius $a=4.0\text{cm}$ carries a uniform volume charge density $\rho =5.0\times 10^{-6}{\text{C/m}}^3$. Determine the magnitude of the electric field (a) at a point $r=2.0\text{cm}$ from the axis (inside the cylinder) and (b) at a point $r=6.0\text{cm}$ from the axis (outside the cylinder). State the direction of the field in each case.

Step-by-step solution

1. Apply Gauss's law: $\oint \mathbf{E}\cdot d\mathbf{A}=\frac{Q_{\text{enc}}}{{\epsilon }_0}$.
2. Choose a cylindrical Gaussian surface of length $L$ coaxial with the charged cylinder.
3. For the inside point ($r<a$):
4. Calculate the charge enclosed: $Q_{\text{enc}}=\rho \pi r^{2}L$.
5. The flux through the curved surface is $E(2\pi rL)$ (field is radial and uniform on the surface).
6. Set up Gauss's law: $E(2\pi rL)=\frac{\rho \pi r^{2}L}{{\epsilon }_0}$.
7. Solve for $E$: $E=\frac{\rho r}{2{\epsilon }_0}$.
8. Insert numbers: $E=\frac{5.0\times 10^{-6}\times 0.02}{2\times 8.85\times 10^{-12}}\approx 5.6\times 10^3\text{N/C}$, directed radially outward.
9. For the outside point ($r>a$):
10. The enclosed charge is the total charge of the cylinder of radius $a$: $Q_{\text{enc}}=\rho \pi a^{2}L$.
11. Flux is again $E(2\pi rL)$.
12. Gauss's law gives: $E(2\pi rL)=\frac{\rho \pi a^{2}L}{{\epsilon }_0}$.
13. Solve for $E$: $E=\frac{\rho a^2}{2{\epsilon }_{0}r}$.
14. Insert numbers: $E=\frac{5.0\times 10^{-6}\times (0.04{)}^2}{2\times 8.85\times 10^{-12}\times 0.06}\approx 7.5\times 10^3\text{N/C}$, also radially outward.
15. Thus the field points away from the axis in both regions.

Concept tested: Gauss's law for volume charge distribution

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