Gauss's law

A fence round a charge

Put a positive charge in the middle of a field. Lines start on it and spread out in all directions. Now build a fence round the charge. The fence can be a ball, a box or a lumpy bag. Every line has to get out, so every line crosses the fence.

The number of lines that cross the fence is fixed by the charge. It does not depend on the shape or the size of the fence. A bigger fence is crossed by the same lines, in a wider spread. This is the idea behind Gauss's law.

The flux of a point charge through a sphere

Let us check the idea for the simplest fence, a sphere with the charge at its centre. At a distance $r$ the field of a point charge $q$ is

The constant $\varepsilon_0$ (epsilon zero) is the permittivity of free space, $8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)$. In Coulomb's law it appears as $\dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9\ \text{N m}^2/\text{C}^2$.

The field is radial and has the same size at every point of the sphere. So the flux is the field times the area of the sphere, $4\pi r^2$:

Look at what happened to $r$. It cancelled. The field falls as $1/r^2$ and the area grows as $r^2$, so the flux is the same for a small sphere and for a huge one. This cancelling is the whole reason Gauss's law works. If the force fell as $1/r^3$, the flux would shrink as the sphere grew, and the law would fail.

Any shape

Why does the shape not matter? Look at the lines. A line that leaves the sphere does not stop, since lines start and end only on charges. It goes on and crosses any other closed surface that surrounds the charge. So the number of lines through the second surface is the same as through the sphere.

We count a line as positive where it leaves and negative where it enters. A line that goes out, comes back in and goes out again crosses the surface three times, but its count is $+1 - 1 + 1 = +1$. So each line from the charge counts once, whatever the shape of the surface.

Gauss's law

Putting the results together, we have Gauss's law:

In words: the net flux through a closed surface equals the net charge inside the surface divided by $\varepsilon_0$.

• $q_{\text{in}}$ is the sum of the charges inside, with their signs.
• A positive $q_{\text{in}}$ gives a net flux that goes outward. A negative $q_{\text{in}}$ gives a net flux that goes inward.
• The surface we choose is imaginary. We call it a Gaussian surface. We can draw it anywhere, and in any shape.

A worked example. A closed surface holds charges of $+5.0$ nC, $-2.2$ nC and $+1.2$ nC. Find the net flux.

1. The charge inside. $q_{\text{in}} = 5.0 - 2.2 + 1.2 = 4.0$ nC $= 4.0 \times 10^{-9}$ C.
2. The flux. $\Phi = \dfrac{q_{\text{in}}}{\varepsilon_0} = \dfrac{4.0 \times 10^{-9}\ \text{C}}{8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 452\ \text{N m}^2/\text{C}$. It is outward, since $q_{\text{in}}$ is positive.

Charges outside the surface

A charge that sits outside the surface sends lines through it. Each line goes in at one place and comes out at another. The positive flux where it leaves cancels the negative flux where it enters, so the net flux from that charge is zero. The right picture above shows this.

Take care here. Gauss's law gives the net flux, not the field. The field on the surface comes from all the charges, those inside and those outside. A charge outside changes the field at the surface. It does not change the net flux.

Think of a box with no charge inside, near a charge that is outside. The net flux through the box is zero. The field at the surface of the box is not zero.

A worked example. A point charge of $9.0$ nC sits at the centre of a cube of side $0.20$ m. Find the flux through one face.

1. The whole cube. $\Phi = \dfrac{q}{\varepsilon_0} = \dfrac{9.0 \times 10^{-9}\ \text{C}}{8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 1017\ \text{N m}^2/\text{C}$.
2. One face. The six faces are alike, so each gets a sixth: $\dfrac{1017}{6} = 169.5\ \text{N m}^2/\text{C}$.

The side of the cube never entered the calculation. A cube ten times larger gives the same answer.

What if the charge moves? Push the charge off the centre, but keep it inside the cube. The flux through the whole cube stays $1017\ \text{N m}^2/\text{C}$. The faces no longer get equal shares. The faces that are closer get more. Now take the charge outside the cube. The net flux through the cube falls to zero.

Gauss's law and Coulomb's law

We built Gauss's law from Coulomb's law. We can go the other way. Put a charge $q$ at the centre of a sphere of radius $r$. Gauss's law gives $E \times 4\pi r^2 = q/\varepsilon_0$. So $E = \dfrac{q}{4\pi\varepsilon_0 r^2}$, the field of a point charge. The force on a second charge $q_0$ there is $qq_0/(4\pi\varepsilon_0 r^2)$, which is Coulomb's law.

The two laws say the same thing about a point charge. The gain of Gauss's law is that it holds for any arrangement of charges, and for surfaces of any shape. In the next lesson we use it to find fields that would be hard to find with Coulomb's law.

Make $q_2$ negative and slide the radius past it. What happens to the flux? Then set $q_1$ and $q_2$ to opposite values with the same size. What is the net flux when both are inside?

Practise with new numbers

Try it in Python

Gauss's law is a statement about a sum over a surface, so a computer can test it. The program adds $\vec{E}\cdot d\vec{S}$ over thousands of small patches on the six faces of a cube. Then it compares the sum with $q/\varepsilon_0$.

Common mistakes

• Counting the charges outside the surface. They add nothing to the net flux. Only $q_{\text{in}}$ counts.
• Thinking the field is zero when the net flux is zero. The flux can be zero while the field is large at every point of the surface.
• Thinking the flux depends on the size or shape of the surface. It depends only on the charge inside.
• Using only the enclosed charges to find the field at the surface. The field comes from all the charges. Gauss's law gives the net flux, and only a symmetric arrangement lets it give the field.
• Forgetting the sign of the charge. A negative $q_{\text{in}}$ gives a negative flux, with the net flux going inward.

Remember these

• Gauss's law: $\oint \vec{E}\cdot d\vec{S} = \dfrac{q_{\text{in}}}{\varepsilon_0}$, with $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)$.
• $q_{\text{in}}$ is the net charge inside the closed surface, with signs. Charges outside add nothing to the net flux.
• The flux does not depend on the shape or the size of the surface.
• A point charge at the centre of a cube sends $\dfrac{q}{6\varepsilon_0}$ through each face.
• Gauss's law holds for every field, and it follows from the inverse square law.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The net flux through a closed surface is $-300\ \text{N m}^2/\text{C}$. What can you say about the charge inside?

2. A charge sits at the centre of a sphere of radius $10$ cm. The sphere is replaced by one of radius $20$ cm. What happens to the net flux, and what happens to the field on the surface?

3. A point charge is just outside a closed box. Is the net flux through the box zero? Is the field zero on its surface?

4. A closed surface holds a charge $+q$ and a charge $-q$. What is the net flux? Is the field zero on the surface?

5. A point charge sits at the centre of a cube. The flux through one face is $80\ \text{N m}^2/\text{C}$. What is the charge?