Electric flux and Gauss's law: applications to insulators

Last week you took long integrals for the ring and the disk. Where there is symmetry, Gauss’s law reduces all of those integrals to two lines. The exam has three types of question: compute the flux through a given surface, find the net flux through a closed surface from the enclosed charge, and derive the field of a sphere, a line or a sheet with Gauss’s law. The key to all three is the same: flux is the amount of field lines passing through a surface, and the net number of lines leaving a closed surface depends on the charge inside.

Golden rules:

1. The angle is always measured from the normal. If the question says “30° with the surface”, it is 60° with the normal.
2. The net flux depends only on the charge inside. However large a charge is outside, it contributes zero to the net flux.
3. Choose the Gaussian surface to match the symmetry: a sphere for a sphere, a cylinder for a line, a box (pillbox) for a sheet.
4. The field inside an insulator is not zero. Count only the charge within $r$.

Electric flux

A uniform field passes through a flat surface of area $A$. The flux is:

$\Phi_E = \vec E \cdot \vec A = EA\cos\theta \qquad [\text{N·m}^2/\text{C}]$

$\vec A$ is the area vector: its magnitude is the area of the surface and its direction is perpendicular (normal) to the surface. $\theta$ is the angle between $\vec E$ and $\vec A$.

If the field is not uniform or the surface is curved, you divide the surface into small pieces and add:

$\Phi_E = \int \vec E \cdot d\vec A$

Example 1. A uniform field $E = 500$ N/C passes through a 0.20 m × 0.30 m rectangle, and the field makes 30° with the normal.

$\Phi = (500)(0.060)\cos 30^\circ = 26.0\ \text{N·m}^2/\text{C}$

Closed surface

On a closed surface (sphere, cube, box) the normal always points outward. So lines going out give positive flux and lines coming in give negative flux. A line that comes from outside and passes through enters once and leaves once: its net contribution is zero. The only lines that contribute to the net flux are those that start or end inside, that is, the charge inside.

Gauss’s law

$\Phi_E = \oint \vec E \cdot d\vec A = \frac{q_{in}}{\varepsilon_0}$

$q_{in}$ is the net charge inside the closed surface (with its sign). $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N·m}^2)$.

Why is it true? Wrap a point charge $q$ in a sphere of radius $r$. The field is perpendicular to the surface everywhere and has magnitude $kq/r^2$. The flux is $(kq/r^2)(4\pi r^2) = 4\pi k q = q/\varepsilon_0$. The $r$ cancelled: the size of the sphere does not matter. If you deform the shape, the same lines pass through and the result stays the same.

Example 2. A closed surface contains $+3.0$ nC and $-5.0$ nC, and there is $+1.0$ nC outside it.

$\Phi = \frac{(+3.0 - 5.0) \times 10^{-9}}{8.85 \times 10^{-12}} = -226\ \text{N·m}^2/\text{C}$

The $+1.0$ nC outside does not enter the calculation. But it changes the field on the surface. Gauss’s law tells you the net flux, not the field at each point.

PITFALL · “If the flux is zero, the field is zero.” Wrong. For an empty cube placed in a uniform field, the net flux is zero but the field is not 0 on every face. What enters through one face leaves through the opposite face.

Example 3 (flux by symmetry). A charge $q = 2.0$ nC sits exactly at the center of a cube of side 0.20 m. What is the flux through one face?

The six faces are symmetric, each takes an equal share:

$\Phi_{\text{face}} = \frac{q}{6\varepsilon_0} = \frac{2.0 \times 10^{-9}}{6(8.85 \times 10^{-12})} = 37.7\ \text{N·m}^2/\text{C}$

On one face the field is neither uniform nor perpendicular. The direct integral is hard; symmetry gives it to you for free.

Finding the field with Gauss’s law

Gauss’s law is always true, but to extract the field from it you need to be able to pull $E$ out of the integral. For that you choose the Gaussian surface so that on every part of the surface:

• $\vec E$ is perpendicular to the surface and has constant magnitude (then $\oint E\,dA = EA$), or
• $\vec E$ is parallel to the surface (then the flux through that part is 0).

This is possible only for three symmetries:

Uniformly charged insulating sphere

Radius $R$, total charge $Q$, volume density $\rho = Q/(\tfrac{4}{3}\pi R^3)$.

Outside ($r > R$): All the charge is inside the Gaussian sphere.

$E(4\pi r^2) = \frac{Q}{\varepsilon_0} \Rightarrow E = \frac{k_e Q}{r^2}$

From outside, the sphere looks like a point charge at its center.

Inside ($r < R$): Only the charge inside the sphere of radius $r$ counts:

$q_{in} = \rho\cdot\tfrac{4}{3}\pi r^3 = Q\frac{r^3}{R^3} \quad\Rightarrow\quad E = \frac{k_e Q}{R^3}\,r$

It is zero at the center, grows linearly with $r$, reaches its largest value $kQ/R^2$ at the surface, then falls as $1/r^2$.

Example 4. $R = 0.10$ m, $Q = +6.0\ \mu$C uniformly distributed.

r = 0.05 m (inside):   E = kQr/R³ = (8.99e9)(6.0e-6)(0.05)/(0.10)³ = 2.70e6 N/C
r = 0.10 m (surface):  E = kQ/R²  = (8.99e9)(6.0e-6)/(0.10)²    = 5.39e6 N/C
r = 0.20 m (outside):  E = kQ/r²  = (8.99e9)(6.0e-6)/(0.20)²    = 1.35e6 N/C

Halfway inside, the field is half of the surface value. At twice the distance outside, it is a quarter.

Infinite line of charge

Charge density $\lambda$. The field points radially away from the line (if positive). Choose a cylinder of radius $r$ and length $\ell$. On the end caps the field is parallel to the surface: flux 0. On the side surface the field is perpendicular and constant:

$E(2\pi r\ell) = \frac{\lambda\ell}{\varepsilon_0} \Rightarrow E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{2k_e\lambda}{r}$

$\ell$ cancelled, as it should. The field falls as $1/r$, not $1/r^2$.

Example 5. $\lambda = 4.0\ \mu\text{C/m}$, $r = 0.12$ m.

$E = \frac{2(8.99 \times 10^{9})(4.0 \times 10^{-6})}{0.12} = 5.99 \times 10^{5}\ \text{N/C}$

The same method applied inside a uniformly charged cylinder of radius $R$ ($q_{in} = \rho\pi r^2\ell$) gives $E = \rho r/(2\varepsilon_0)$.

Infinite insulating sheet

Surface density $\sigma$. Choose a box (pillbox) of cross-section area $A$ that passes through the sheet. On the side faces the field is parallel: flux 0. On the two end caps the field is perpendicular and outward:

$E(2A) = \frac{\sigma A}{\varepsilon_0} \Rightarrow E = \frac{\sigma}{2\varepsilon_0}$

This is the same result we found last week from the $R \to \infty$ limit of the disk formula, but without an integral. For two oppositely charged parallel sheets, the fields add in between ($\sigma/\varepsilon_0$) and cancel outside (0). This is the foundation of the capacitor topic.

When does Gauss’s law not help?

For a finite rod, a ring or a disk, Gauss’s law is still true, but there is no surface on which $E$ is both constant and perpendicular. You can’t pull $E$ out of the integral. That is why last week’s integral method was needed. If you see a question like “use Gauss’s law to find the field of a finite rod” on the exam, there is a catch: either the rod is to be treated as “very long” or you go back to the integral method. The rule is simple: first show the symmetry, then draw the Gaussian surface.

TIP · Check the result by dimension. Point: $1/r^2$. Line: $1/r$. Plane: independent of $r$. As the number of dimensions of the source grows, the field falls off more slowly with distance.

Common mistakes