Gauss's Law & Applications

Step 1: Identify the symmetry

The charge distribution has spherical symmetry. Therefore $\mathbf{E}$ must be radial and depend only on $r$: $\mathbf{E} = E_r(r)\,\hat{r}$. Choose a concentric spherical Gaussian surface of radius $r$.

Step 2: Evaluate the flux integral

On the Gaussian sphere, $\mathbf{E}$ is constant in magnitude and parallel to $d\mathbf{a} = r^2\sin\theta\,d\theta\,d\phi\,\hat{r}$:

$$\oint_S \mathbf{E}\cdot d\mathbf{a} = E_r(r)\oint_S da = E_r(r)\cdot 4\pi r^2$$

Step 3: Case I — Outside the sphere ($r > R$)

The entire charge $Q$ is enclosed: $Q_{\rm enc} = Q$. Applying Gauss's law:

$$E_r(r)\cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \quad\Longrightarrow\quad \boxed{E_r = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}} \quad (r > R)$$

The sphere behaves exactly like a point charge at its center — this is the shell theorem.

Step 4: Case II — Inside the sphere ($r < R$)

Only the charge within radius $r$ is enclosed. Since $\rho$ is uniform:

$$Q_{\rm enc} = \rho\cdot\frac{4}{3}\pi r^3 = Q\frac{r^3}{R^3}$$

Step 5: Apply Gauss's law inside

$$E_r(r)\cdot 4\pi r^2 = \frac{Q}{\epsilon_0}\frac{r^3}{R^3} \quad\Longrightarrow\quad \boxed{E_r = \frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}} \quad (r < R)$$

The field grows linearly with $r$ inside the sphere, reaching its maximum at $r = R$.

Step 6: Verify continuity at $r = R$

From outside: $E_r(R) = Q/(4\pi\epsilon_0 R^2)$. From inside: $E_r(R) = QR/(4\pi\epsilon_0 R^3) = Q/(4\pi\epsilon_0 R^2)$. The field is continuous at the surface, as expected for a volume charge distribution (no surface charge discontinuity).

Step 1: Identify the symmetry

By planar symmetry, $\mathbf{E}$ must be perpendicular to the plane and can only depend on the distance from it. Let the plane be at $z = 0$. Then $\mathbf{E} = E_z(z)\,\hat{z}$, and by the symmetry $z \to -z$: $E_z(-z) = -E_z(z)$.

Step 2: Construct the Gaussian pillbox

Choose a cylindrical pillbox with cross-sectional area $A$, straddling the plane symmetrically — one cap at $z = +h$, the other at $z = -h$.

Step 3: Evaluate the flux through each surface

The pillbox has three surfaces: top cap, bottom cap, and curved side. By symmetry, $\mathbf{E}$ is parallel to the curved wall (perpendicular to its outward normal), so the curved surface contributes zero flux:

$$\Phi_{\rm curved} = 0$$

On the top cap ($z = +h$), $\mathbf{E}$ points upward and $d\mathbf{a} = da\,\hat{z}$: flux $= E_z A$. On the bottom cap ($z = -h$), $\mathbf{E}$ points downward and $d\mathbf{a} = da\,(-\hat{z})$: flux $= (-E_z)(-A) = E_z A$.

Step 4: Compute $Q_{\rm enc}$ and apply Gauss's law

The enclosed charge is $Q_{\rm enc} = \sigma A$. Gauss's law gives:

$$2E_z A = \frac{\sigma A}{\epsilon_0}$$

Step 5: Solve for $\mathbf{E}$

$$\boxed{\mathbf{E} = \frac{\sigma}{2\epsilon_0}\,\hat{n}}$$

where $\hat{n}$ points away from the surface on each side. The field is uniform — independent of distance from the plane. This is consistent with our Coulomb derivation for a disk in the limit $R\to\infty$.

Step 1: Set up a thin Gaussian pillbox at the surface

Consider a surface carrying charge density $\sigma$. Construct a tiny pillbox of area $A$ and thickness $\varepsilon \to 0$, straddling the surface. Let $\hat{n}$ point from "below" to "above."

Step 2: Apply Gauss's law to the normal component

As $\varepsilon\to 0$, the curved side contributes zero flux. The top and bottom caps give:

$$E_{\rm above}^\perp A - E_{\rm below}^\perp A = \frac{\sigma A}{\epsilon_0}$$

Step 3: Normal component discontinuity

Dividing by $A$:

$$\boxed{E_{\rm above}^\perp - E_{\rm below}^\perp = \frac{\sigma}{\epsilon_0}}$$

The normal component of $\mathbf{E}$ is discontinuous by $\sigma/\epsilon_0$ across any surface charge layer.

Step 4: Apply Faraday's law to the tangential component

Since $\nabla\times\mathbf{E} = 0$ in electrostatics, construct a thin rectangular loop straddling the surface, with sides of length $l$ parallel to the surface and width $\varepsilon\to 0$. Stokes' theorem gives:

$$\oint \mathbf{E}\cdot d\boldsymbol{\ell} = 0$$

Step 5: Tangential component continuity

As $\varepsilon\to 0$, the short sides contribute nothing. The two long sides give:

$$E_{\rm above}^\parallel\,l - E_{\rm below}^\parallel\,l = 0$$

$$\boxed{E_{\rm above}^\parallel = E_{\rm below}^\parallel}$$

The tangential component of $\mathbf{E}$ is always continuous across a surface charge — there is no tangential discontinuity in electrostatics.

Step 6: Combined vector boundary condition

The full boundary condition can be written compactly as:

$$\boxed{\mathbf{E}_{\rm above} - \mathbf{E}_{\rm below} = \frac{\sigma}{\epsilon_0}\,\hat{n}}$$

The discontinuity is entirely in the normal direction, with magnitude $\sigma/\epsilon_0$.