The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Chapter 22 — Gauss's Law

Problems with worked solutions · 12 points · 3 extra credit

Chapter 22 — Gauss's Law

Core: 12 points · Extra credit: 3 points. Show your reasoning; partial credit is given. Use $\varepsilon_0=8.85\times10^{-12}\,\text{C}^2/(\text{N}\cdot\text{m}^2)$ and $k=8.99\times10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2$.

1. 1. (2 pts ••) A uniform electric field $E=6.0\times10^{3}\,\text{N/C}$ points in the $+x$ direction.

   (a) Find the electric flux through a flat $0.20\,\text{m}\times0.40\,\text{m}$ rectangle whose normal makes a $30^\circ$ angle with the field.

   (b) A closed box encloses a net charge of $+8.0\,\text{nC}$. What is the net electric flux through the box?

   Solution

   1.

   (a) $\Phi=EA\cos\theta=(6.0\times10^{3})(0.20\times0.40)\cos30^\circ=(6.0\times10^{3})(0.080)(0.866)\approx 4.2\times10^{2}\,\text{N}\cdot\text{m}^2/\text{C}$.

   (b) $\Phi_\text{net}=\dfrac{q_\text{enc}}{\varepsilon_0}=\dfrac{8.0\times10^{-9}}{8.85\times10^{-12}}\approx 9.0\times10^{2}\,\text{N}\cdot\text{m}^2/\text{C}$.

2. 2. (2 pts ••) A point charge $q=+5.0\,\text{nC}$ sits at the center of a cube.

   (a) Find the total electric flux through the cube.

   (b) Find the flux through one face, and explain how symmetry gives this in one more step.

   Solution

   2.

   (a) $\Phi=\dfrac{q}{\varepsilon_0}=\dfrac{5.0\times10^{-9}}{8.85\times10^{-12}}\approx 5.6\times10^{2}\,\text{N}\cdot\text{m}^2/\text{C}$.

   (b) The charge sits at the center, so by symmetry the six faces share the flux equally: $\Phi_\text{face}=\dfrac{\Phi}{6}\approx 94\,\text{N}\cdot\text{m}^2/\text{C}$.

3. 3. (2 pts ••) A solid insulating sphere of radius $R=0.10\,\text{m}$ carries charge $Q=+4.0\,\text{nC}$ spread uniformly through its volume.

   (a) Find the electric field at $r=0.20\,\text{m}$.

   (b) Find the electric field at $r=0.050\,\text{m}$.

   Solution

   3.

   (a) Outside ($r>R$) the sphere acts like a point charge: $E=\dfrac{kQ}{r^2}=\dfrac{(8.99\times10^{9})(4.0\times10^{-9})}{(0.20)^2}\approx 9.0\times10^{2}\,\text{N/C}$.

   (b) Inside ($r<R$) only the enclosed charge counts; for uniform density $E=\dfrac{kQr}{R^3}=\dfrac{(8.99\times10^{9})(4.0\times10^{-9})(0.050)}{(0.10)^3}\approx 1.8\times10^{3}\,\text{N/C}$.

4. 4. (2 pts ••) An infinitely long line carries uniform linear charge density $\lambda=3.0\,\text{nC/m}$. Using a Gaussian cylinder, find the magnitude and direction of the electric field at a distance $r=5.0\,\text{cm}$ from the line.

   Solution

   4. A coaxial Gaussian cylinder of radius $r$ and length $\ell$ encloses $\lambda\ell$; only the curved wall has flux, so $E(2\pi r\ell)=\dfrac{\lambda\ell}{\varepsilon_0}$, giving $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}=\dfrac{2k\lambda}{r}=\dfrac{2(8.99\times10^{9})(3.0\times10^{-9})}{0.050}\approx 1.1\times10^{3}\,\text{N/C}$, directed radially outward.

5. 5. (3 pts •••) A point charge $q=+2.0\,\text{nC}$ sits at the center of a neutral conducting spherical shell of inner radius $a$ and outer radius $b$.

   (a) Find the charge induced on the inner and outer surfaces of the shell.

   (b) Find the electric field in the three regions $r<a$, $a<r<b$, and $r>b$.

   (c) How do your answers to (a) and (b) change if the shell instead carries a net charge of $-5.0\,\text{nC}$?

   Solution

   5.

   (a) The field must vanish inside the conductor, so the inner surface carries $-2.0\,\text{nC}$ (canceling the enclosed charge). The shell is neutral, so the outer surface carries $+2.0\,\text{nC}$.

   (b) $r<a$: $E=\dfrac{kq}{r^2}$ (from the central charge). $a<r<b$: $E=0$ (inside the conductor). $r>b$: $E=\dfrac{kq}{r^2}$, from the $+2.0\,\text{nC}$ on the outer surface.

   (c) The inner surface is still $-2.0\,\text{nC}$ (it must cancel the enclosed $+2.0\,\text{nC}$). The outer surface now carries $-5.0-(-2.0)=-3.0\,\text{nC}$. The fields for $r<a$ and $a<r<b$ are unchanged; for $r>b$ the enclosed total is $2.0-5.0=-3.0\,\text{nC}$, so $E=\dfrac{k(3.0\,\text{nC})}{r^2}$ directed radially inward.

6. 6. (1 pt •) A closed surface encloses a $+3.0\,\text{nC}$ charge and a $-5.0\,\text{nC}$ charge and no others. What is the net electric flux through the surface?

   Solution

   6. $\Phi=\dfrac{q_\text{enc}}{\varepsilon_0}=\dfrac{(3.0-5.0)\times10^{-9}}{8.85\times10^{-12}}\approx -2.3\times10^{2}\,\text{N}\cdot\text{m}^2/\text{C}$ (net inward).

7. Extra Credit (3 pts •••)

   A solid insulating sphere of radius $R$ has a radially varying charge density $\rho(r)=\rho_0\!\left(1-\dfrac{r}{R}\right)$ for $r\le R$ (and zero outside).

   (a) Find the total charge $Q$ of the sphere.

   (b) Use Gauss's law to find the electric field $E(r)$ both inside ($r<R$) and outside ($r>R$).

   (c) Find the radius at which the field inside the sphere is maximum.

   Solution

   Extra Credit

   (a) $Q=\displaystyle\int_0^R \rho_0\!\left(1-\frac{r}{R}\right)4\pi r^2\,dr=4\pi\rho_0\left[\frac{R^3}{3}-\frac{R^3}{4}\right]=\frac{\pi\rho_0 R^3}{3}.$

   (b) For $r<R$ the enclosed charge is

   $$q_\text{enc}(r)=\int_0^r \rho_0\!\left(1-\frac{r'}{R}\right)4\pi r'^2\,dr'=4\pi\rho_0\left(\frac{r^3}{3}-\frac{r^4}{4R}\right),$$

   so Gauss's law $E\,(4\pi r^2)=q_\text{enc}/\varepsilon_0$ gives

   $$E(r)=\frac{\rho_0}{\varepsilon_0}\left(\frac{r}{3}-\frac{r^2}{4R}\right)\quad (r<R).$$

   For $r>R$, all of $Q$ is enclosed: $E(r)=\dfrac{Q}{4\pi\varepsilon_0 r^2}=\dfrac{\rho_0 R^3}{12\,\varepsilon_0 r^2}.$

   (c) Set $\dfrac{dE}{dr}=\dfrac{\rho_0}{\varepsilon_0}\left(\frac13-\frac{r}{2R}\right)=0\Rightarrow r=\dfrac{2R}{3}.$