Electric flux

Rain through a hoop

Hold a hoop in the rain. Face it straight into the rain and many drops pass through. Tilt it, and fewer drops pass. Hold it edge on, and no drop passes at all.

The electric field behaves in a similar way. Think of the field lines as rain: more lines where the field is stronger. The number of lines that pass through a surface is called the electric flux. It tells us how much of the field crosses the surface.

A warning first. Nothing flows along a field line. The lines are a drawing, and the flux is a count made with that drawing. We use the rain only to see how the count depends on the size and the tilt of the surface.

The flux through a flat surface

Take a flat surface of area $A$ in a uniform field $E$. Three cases tell us nearly everything.

• The surface faces the field. The most lines cross it.
• The surface is turned a little. Fewer lines cross it.
• The surface lies along the field. No line crosses it.

To describe the tilt, we draw an arrow of length $A$ at right angles to the surface. This arrow is the area vector. It points along the normal to the surface. The angle $\theta$ is the angle between the field and the normal.

Look at the middle picture. A surface turned by $\theta$ catches only the lines that cross its shadow, and the shadow has the area $A\cos\theta$. So the flux is

The Greek letter $\Phi$ (phi) is the symbol for flux. In vectors this is the dot product $\Phi = \vec{E}\cdot\vec{A}$.

The unit follows from the formula. The field is in N/C and the area is in m², so flux is in $\text{N m}^2/\text{C}$. A volt is a newton metre for each coulomb, so the same unit can be written as a volt metre, V m.

Flux is a scalar. It has a size and a sign, but no direction.

• At $\theta = 0^\circ$ the flux is $+EA$, the largest value.
• At $\theta = 90^\circ$ the flux is zero. The surface lies along the field.
• At $\theta = 180^\circ$ the flux is $-EA$. The field now passes through from the other side.

A worked example. A square frame of side $0.20$ m is in a uniform field of $600$ N/C. Find the flux when (a) the frame faces the field, (b) its normal makes $60^\circ$ with the field, (c) the frame lies along the field.

1. The area. $A = (0.20\ \text{m})^2 = 0.040\ \text{m}^2$.
2. Facing the field. $\theta = 0^\circ$, so $\Phi = EA = 600\ \text{N/C} \times 0.040\ \text{m}^2 = 24\ \text{N m}^2/\text{C}$.
3. Turned by 60°. $\Phi = EA\cos 60^\circ = 24 \times 0.50 = 12\ \text{N m}^2/\text{C}$. That is half of the largest flux, as the picture shows.
4. Along the field. The normal is at $90^\circ$ to the field, so $\Phi = 0$.

Which angle? The angle in the formula is always the angle with the normal. Sometimes a problem gives the angle between the field and the plane of the surface. Then the angle with the normal is $90^\circ$ minus that angle. A field that makes $25^\circ$ with a plane makes $65^\circ$ with its normal, and the flux is $EA\cos 65^\circ$, or $EA\sin 25^\circ$.

Set the angle to $90^\circ$. Then slide it up to $180^\circ$. What happens to the sign of the flux, and why?

Which side is outward?

A closed surface, such as a ball or a box, has an inside and an outside. For a closed surface we always draw the normal pointing outward. This fixes the sign.

• Where lines leave the surface, the angle between the field and the outward normal is less than $90^\circ$. The flux is positive.
• Where lines enter the surface, the angle is more than $90^\circ$. The flux is negative.

The net flux through a closed surface is the sum of the flux through all its parts. It tells us how many more lines leave than enter.

A worked example. A closed tin can has a radius of $5.0$ cm. It lies with its axis along a uniform field of $300$ N/C. Find the flux through each end, through the curved side, and the net flux.

1. The area of an end. $A = \pi r^2 = \pi \times (0.050\ \text{m})^2 = 7.85 \times 10^{-3}\ \text{m}^2$.
2. The end that the field enters. The outward normal points against the field, so $\theta = 180^\circ$. $\Phi = -EA = -300\ \text{N/C} \times 7.85 \times 10^{-3}\ \text{m}^2 = -2.36\ \text{N m}^2/\text{C}$.
3. The end that the field leaves. Here $\theta = 0^\circ$, so $\Phi = +2.36\ \text{N m}^2/\text{C}$.
4. The curved side. The field runs along it, so the normal is at $90^\circ$ to the field. The flux is zero.
5. The net flux. $-2.36 + 2.36 + 0 = 0$.

The net flux is zero. Every line that enters on one end leaves on the other. This is true for any closed surface in a uniform field. A bottle, a box and a ball all give zero, because the lines that go in come out again.

A field that changes from place to place

Most fields are not uniform. Then we cut the surface into small patches. Each patch is so small that it is nearly flat and the field on it is nearly constant. For one patch with area $\Delta S$,

We add the flux of all the patches. For a closed surface this sum is written $\oint \vec{E}\cdot d\vec{S}$, and the circle on the integral sign reminds us that the surface is closed.

There is one case where the sum is easy. Put a point charge $q$ at the centre of a sphere of radius $r$. By symmetry, the field has the same size at every point of the sphere. It points straight out, along the outward normal, so $\theta = 0^\circ$ for every patch. Then the sum is simply $E$ times the whole area, $E \times 4\pi r^2$.

For this sphere the net flux is not zero. Lines start on the charge and all of them leave. So a closed surface can have a net flux when it holds charge. The next lesson shows exactly how much.

Practise with new numbers

Try it in Python

The program finds the flux through each of the six faces of a box that sits in a uniform field. The field points partly along $x$ and partly along $y$.

Common mistakes

• Using the angle with the plane. The cosine needs the angle with the normal. If the angle is with the plane, use the sine.
• Thinking flux is a vector. It is a scalar. It can be positive, negative or zero, and it has no direction.
• Forgetting the sign on a closed surface. Lines that enter count as negative. That is why the net flux of a box in a uniform field is zero.
• Saying that zero flux means zero field. A surface along the field has zero flux, but the field is not zero.
• Mixing cm² and m². $1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$.

Remember these

• $\Phi = EA\cos\theta = \vec{E}\cdot\vec{A}$, with $\theta$ the angle between the field and the normal. The unit is $\text{N m}^2/\text{C}$.
• Facing the field: $+EA$. Along the field: $0$. Facing away: $-EA$.
• For a closed surface the normal points outward. Lines that leave give positive flux, lines that enter give negative flux.
• In a uniform field the net flux through any closed surface is zero.
• In a field that changes, add the flux of small patches: $\Phi = \sum \vec{E}\cdot\Delta\vec{S}$.

Check yourself

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

1. A flat coil of area $0.050\ \text{m}^2$ faces a uniform field of $240$ N/C. What is the flux? What is the flux after the coil turns through $90^\circ$ about an axis in its own plane?

2. A surface has an area of $0.10\ \text{m}^2$. The field makes $30^\circ$ with the plane of the surface. Is the flux $EA\cos 30^\circ$ or $EA\sin 30^\circ$? Which fraction of $EA$ is it?

3. Why is the net flux through a closed box zero when the box sits in a uniform field?

4. The net flux through a closed surface is not zero. What does this tell you about the field and about what is inside?