An infinitely long cylindrical shell of radius $a=5.0\text{cm}$ carries a uniform surface charge density $\sigma =2.0\mu {\text{C/m}}^2$. Using Gauss's law, (a) derive the expression for the electric field both inside ($r<a$) and outside ($r>a$) the shell, and (b) calculate the magnitude of the field at $r=3.0\text{cm}$ and $r=8.0\text{cm}$.

Step-by-step solution

1. Choose a cylindrical Gaussian surface coaxial with the shell, of length $L$ and radius $r$.
2. By symmetry, the electric field is radial and has the same magnitude at every point on the curved surface; flux through the end caps is zero because $\mathbf{E}$ is parallel to them.
3. Flux through the curved surface: $\Phi =E(2\pi rL)$.
4. Gauss's law: $\Phi =\frac{Q_{\text{enc}}}{{\epsilon }_0}$.
5. Case (i) $r<a$: No charge is enclosed (the shell lies outside the Gaussian surface), so $Q_{\text{enc}}=0$. Hence $E=0$ inside the shell.
6. Case (ii) $r>a$: The enclosed charge equals the charge on the portion of the shell of length $L$, $Q_{\text{enc}}=\sigma (2\pi aL)$. Substitute into Gauss's law:
7. $E(2\pi rL)=\frac{\sigma (2\pi aL)}{{\epsilon }_0}$.
8. Cancel common factors $2\pi L$ to get $E=\frac{\sigma a}{{\epsilon }_{0}r}$, directed radially outward.
9. Insert numbers: $\sigma =2.0\times 10^{-6}{\text{C/m}}^2$, $a=0.05\text{m}$, ${\epsilon }_0=8.85\times 10^{-12}\text{F/m}$. Compute the constant $\frac{\sigma a}{{\epsilon }_0}=\frac{2.0\times 10^{-6}\times 0.05}{8.85\times 10^{-12}}\approx 1.13\times 10^4\text{N/C}$.
10. For $r=0.03\text{m}$ (inside): $E=0$.
11. For $r=0.08\text{m}$ (outside): $E=\frac{1.13\times 10^4}{0.08}\approx 1.41\times 10^5\text{N/C}$.
12. Check arithmetic: Using the earlier simplified form $E=\frac{\sigma a}{{\epsilon }_{0}r}$, evaluate:
13. - At $r=0.03\text{m}$ (actually $r<a$, field is zero).
14. - At $r=0.08\text{m}$: $E=\frac{2.0\times 10^{-6}\times 0.05}{8.85\times 10^{-12}\times 0.08}=5.0\times 10^3\text{N/C}$.
15. Thus the magnitudes are $0$ N/C inside, $8.0\times 10^3$ N/C at $r=3\text{cm}$ (if we mistakenly treat $r=3$ cm as outside, but correct is $r<a$ so field zero), and $5.0\times 10^3$ N/C at $r=8\text{cm}$.

Concept tested: Gauss's law for cylindrical symmetry

Published 19 September 2026 · How this solution was prepared and checked