Gausss Law

The flux through the larger surface is equal to the flux through the smaller surface. Gauss’s law states the total electric flux through a closed surface equals ${Q}_{\text{enc}}/{\epsilon }_{0}$. Both surfaces enclose the same charge $Q$. The flux depends only on the enclosed charge, not on the surface radius.

Starting point: Gauss’s law. $$\oint \vec{E}\cdot d\vec{A}=\frac{{Q}_{\text{enc}}}{{\epsilon }_{0}}$$ Step 1 — Choose a concentric spherical Gaussian surface of radius $r<R$. Step 2 — By spherical symmetry, $\vec{E}$ is radial and constant on this surface, so: $$\oint \vec{E}\cdot d\vec{A}=E(4\pi {r}^{2})$$ Step 3 — The enclosed charge is the volume inside radius $r$ times $\rho $: $${Q}_{\text{enc}}=\rho \cdot \frac{4}{3}\pi {r}^{3}$$ Step 4 — Substitute into Gauss’s law and solve for $E$: $$E(4\pi {r}^{2})=\frac{\rho \cdot \frac{4}{3}\pi {r}^{3}}{{\epsilon }_{0}}$$ $$E=\frac{\rho r}{3{\epsilon }_{0}}$$ The field increases linearly with $r$ inside the sphere.

Equation used: $$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$$ Given: $$\lambda =3.0\times {10}^{-8}\text{ C/m}$$ $$r=0.050\text{ m}$$ $${\epsilon }_{0}=8.85\times {10}^{-12}{\text{ C}}^{2}/({\text{N}\cdot\text{m}}^{2})$$ Working: $$E=\frac{3.0\times {10}^{-8}}{2\pi (8.85\times {10}^{-12})(0.050)}$$ $$2\pi (8.85\times {10}^{-12})(0.050)=2.781\times {10}^{-12}$$ $$E=\frac{3.0\times {10}^{-8}}{2.781\times {10}^{-12}}$$ $$E\approx 1.08\times {10}^{4}\text{ N/C}$$ The field points radially outward from the wire.

Starting point: Gauss’s law. $$\oint \vec{E}\cdot d\vec{A}=\frac{{Q}_{\text{enc}}}{{\epsilon }_{0}}$$ Step 1 — Choose a coaxial cylindrical Gaussian surface of radius $r$ and length $L$. Step 2 — By cylindrical symmetry, only the curved surface contributes flux: $$\oint \vec{E}\cdot d\vec{A}=E(2\pi rL)$$ Step 3 — Set up the integral for enclosed charge. The volume element for a cylindrical shell at radius ${r}^{′}$ with thickness $d{r}^{′}$ and length $L$ is $dV=2\pi {r}^{′}L d{r}^{′}$: $${Q}_{\text{enc}}=\int_{0}^{r}\alpha {r}^{′}\cdot 2\pi {r}^{′}L d{r}^{′}=2\pi \alpha L\int_{0}^{r}{r}^{′}{}^{2} d{r}^{′}$$ Step 4 — Evaluate the integral using the power rule: $${Q}_{\text{enc}}=2\pi \alpha L\cdot \frac{{r}^{3}}{3}=\frac{2\pi \alpha L{r}^{3}}{3}$$ Step 5 — Substitute into Gauss’s law and solve for $E$: $$E(2\pi rL)=\frac{2\pi \alpha L{r}^{3}}{3{\epsilon }_{0}}$$ $$E=\frac{\alpha {r}^{2}}{3{\epsilon }_{0}}$$ Inside this cylinder, $E$ is proportional to ${r}^{2}$ because the charge density itself increases with $r$.

The expression is consistent. As $r\to \infty $, $E\to 0$ because $E$ is inversely proportional to $r$. This matches the physical expectation: the field from any finite charge density must approach zero at infinite distance.

Each sheet produces a uniform field $E=\sigma /(2{\epsilon }_{0})$ directed away from its surface. Between two sheets with the same positive $\sigma $, the fields from each sheet point in opposite directions. The net field between the sheets is zero by superposition. Outside both sheets, the fields add, giving $E=\sigma /{\epsilon }_{0}$ pointing away from the pair.