Fields from Gauss's law

Choosing the surface

Gauss's law is always true. It gives us the field only when we can pull $E$ out of the sum. That happens when the charges have a symmetry, and we choose a surface that fits it. On each part of the surface, the field must either have the same size and be at right angles to the surface, or run along the surface and give no flux.

• Sphere symmetry (a ball, a shell, a point charge). The field points straight out and depends only on the distance from the centre. Use a sphere round the centre.
• Line symmetry (a very long straight wire). The field points straight out from the wire. Use a cylinder with the wire as its axis.
• Plane symmetry (a large flat sheet). The field is at right angles to the sheet, on both sides. Use a short box that goes through the sheet.

We can find three fields in this way. Then we turn to conductors.

A thin spherical shell

A thin shell of radius $R$ carries a total charge $Q$ spread evenly over its surface.

Outside the shell ($r > R$). Draw a Gaussian sphere of radius $r$ round the centre. The field is radial, with the same size on the whole sphere. So the flux is $E \times 4\pi r^2$. The charge inside is $Q$, so

Outside, the shell acts as if all its charge were at the centre.

Inside the shell ($r < R$). A Gaussian sphere inside the shell holds no charge. So $E \times 4\pi r^2 = 0$, and $E = 0$ everywhere inside.

A worked example. A thin shell of radius $0.20$ m carries $8.0$ nC. Find the field at $0.50$ m, at $0.20$ m just outside, and at $0.10$ m.

1. At 0.50 m. $E = \dfrac{Q}{4\pi\varepsilon_0 r^2} = \dfrac{8.0 \times 10^{-9}\ \text{C}}{4\pi \times 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2) \times (0.50\ \text{m})^2} = 288\ \text{N/C}$.
2. Just outside, at 0.20 m. The same formula with $r = 0.20$ m gives $1798\ \text{N/C}$. Check it with a ratio: the distance is $2.5$ times smaller, so the field is $2.5^2 = 6.25$ times larger. And $288 \times 6.25 = 1800$.
3. At 0.10 m. This is inside the shell, so $E = 0$.

A uniformly charged solid ball

Now fill a ball of radius $R$ with charge $Q$, spread evenly through the volume. This is an insulating ball.

Outside ($r > R$). Nothing changes. $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$.

Inside ($r < R$). A Gaussian sphere of radius $r$ holds only a part of the charge. The volume grows as $r^3$, so the sphere holds the fraction $r^3/R^3$ of $Q$:

Inside the ball the field rises in a straight line from zero at the centre. At the surface it has the value $\dfrac{Q}{4\pi\varepsilon_0 R^2}$. Outside, it falls as $1/r^2$. So the field is largest at the surface of the ball.

A worked example. A solid ball of radius $0.15$ m holds $9.0$ nC spread evenly. Find the field at $0.05$ m and at $0.30$ m from the centre.

1. At 0.05 m, inside. $E = \dfrac{Qr}{4\pi\varepsilon_0 R^3} = \dfrac{9.0 \times 10^{-9}\ \text{C} \times 0.05\ \text{m}}{4\pi \times 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2) \times (0.15\ \text{m})^3} = 1199\ \text{N/C}$.
2. At 0.30 m, outside. $E = \dfrac{Q}{4\pi\varepsilon_0 r^2} = \dfrac{9.0 \times 10^{-9}\ \text{C}}{4\pi \times 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2) \times (0.30\ \text{m})^2} = 899\ \text{N/C}$.

Slide the distance $r$ across the radius $R$. What happens to the field of the shell? What happens to the field of the ball? Where do the two agree?

A long straight wire

A very long wire carries a charge $\lambda$ (lambda) on each metre of its length. By symmetry the field points straight away from the wire, with the same size at every point at distance $r$.

Draw a Gaussian cylinder of radius $r$ and length $l$, with the wire along its axis.

• On the curved side the field is at right angles to the surface, with the same size everywhere. The flux is $E \times 2\pi r l$.
• On the two flat ends the field runs along the surface. The flux is zero.
• The charge inside is $\lambda l$.

So $E \times 2\pi r l = \dfrac{\lambda l}{\varepsilon_0}$, and the length $l$ cancels:

The field falls as $1/r$, more slowly than the field of a point charge. For a negative wire the field points towards the wire. The result is good for points close to the wire, compared with its length, and far from its ends.

A worked example. A long wire carries $4.0$ nC on each metre. Find the field $0.50$ m from it.

$E = \dfrac{\lambda}{2\pi\varepsilon_0 r} = \dfrac{4.0 \times 10^{-9}\ \text{C/m}}{2\pi \times 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2) \times 0.50\ \text{m}} = 144\ \text{N/C}$.

A large flat sheet

A large sheet carries a charge $\sigma$ (sigma) on each square metre. By symmetry the field points straight away from the sheet on both sides, and its size depends only on the distance.

Draw a short box through the sheet, with two flat ends of area $A$, one on each side.

• Through each end the field is at right angles, so the flux is $EA$. The two ends give $2EA$.
• Through the sides the field runs along the surface, so the flux is zero.
• The charge inside is $\sigma A$.

So $2EA = \dfrac{\sigma A}{\varepsilon_0}$, and the area cancels:

Look at what is missing. There is no $r$ in the answer. The field of a large sheet is the same at every distance. It is uniform, because the lines run straight out, and the sheet never looks smaller from far away.

Two opposite sheets. Place two large parallel sheets, one with $+\sigma$ and one with $-\sigma$. Between them, the field of each sheet points the same way, from the positive sheet to the negative one. The two fields add: $E = \dfrac{\sigma}{2\varepsilon_0} + \dfrac{\sigma}{2\varepsilon_0} = \dfrac{\sigma}{\varepsilon_0}$. Outside, the two fields point in opposite directions and cancel. This is the field between the plates of a capacitor.

A worked example. A large sheet carries $8.85$ nC on each square metre. The field is $E = \dfrac{\sigma}{2\varepsilon_0} = \dfrac{8.85 \times 10^{-9}}{2 \times 8.85 \times 10^{-12}} = 500\ \text{N/C}$. Two such sheets with opposite charges give $1000\ \text{N/C}$ between them and zero outside.

Move the distance from $5$ cm to $50$ cm. Which field changes, and which stays the same? At what distance do the two fields meet?

Conductors in electrostatics

A conductor has free charges. When nothing moves, the charges are at rest, and this gives us four facts.

1. The field inside a conductor is zero. If there were a field inside, the free charges would feel a force and move. They would not be at rest.

2. The extra charge sits on the surface. Draw a Gaussian surface inside the metal, just under the surface. The field is zero on it, so the flux is zero, and $q_{\text{in}} = 0$. All the charge must be on the outer surface.

3. Just outside the surface, the field is at right angles to the surface, and $E = \sigma/\varepsilon_0$. Draw a small box that goes through the surface. Its lower end is in the metal, where $E = 0$. Its upper end is in the air and has the area $A$. The sides carry no flux. So $EA = \sigma A/\varepsilon_0$.

The result is twice the field of an insulating sheet. The reason is that the charge on the surface patch itself makes $\sigma/(2\varepsilon_0)$ on each side. The rest of the charge on the conductor makes a field of the same size. Inside the metal the two cancel. Outside they add, and the total is $\sigma/\varepsilon_0$.

4. A cavity is shielded. If a hollow conductor has no charge inside the cavity, the field in the cavity is zero. A Gaussian surface round the cavity, in the metal, shows that the inner surface holds no charge. A sensitive instrument in a metal box does not feel outside fields.

What if there is a charge $+q$ inside the cavity? Then a Gaussian surface in the metal still has zero field on it, so it must enclose zero charge. The inner surface must carry $-q$. If the conductor is neutral, the outer surface carries $+q$, and the field outside is the field of $+q$ at the centre.

A worked example. A metal sphere of radius $0.25$ m carries $10$ nC. Find the surface density of charge and the field just outside.

1. The density. $\sigma = \dfrac{Q}{4\pi R^2} = \dfrac{10 \times 10^{-9}\ \text{C}}{4\pi \times (0.25\ \text{m})^2} = 1.27 \times 10^{-8}\ \text{C/m}^2$.
2. The field. $E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{1.27 \times 10^{-8}\ \text{C/m}^2}{8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 1439\ \text{N/C}$.
3. Check. The shell formula gives $\dfrac{Q}{4\pi\varepsilon_0 R^2}$ at the surface. This is the same number.

Practise with new numbers

Try it in Python

The program gives the fields of a wire, a sheet and a point charge at three distances. It shows how each field changes when the distance doubles.

Common mistakes

• Using Gauss's law for a shape with no symmetry. The law still holds, but it does not give the field when $E$ cannot be pulled out of the sum.
• Using $4\pi$ for a wire. A wire has $2\pi r$ in its denominator. A point charge or a sphere has $4\pi r^2$.
• Forgetting the factor 2 for a sheet. An insulating sheet gives $\sigma/(2\varepsilon_0)$. The field just outside a conductor is $\sigma/\varepsilon_0$.
• Thinking the field is zero outside a charged shell. It is zero inside. Outside it is the field of a point charge.
• Using $r$ in place of $R$ in a ball. Inside a solid ball the field is $Qr/(4\pi\varepsilon_0 R^3)$, with $r$ the distance from the centre.

Remember these

• Pick the Gaussian surface that fits the symmetry: a sphere, a cylinder or a box.
• Thin shell or ball, outside: $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$. Thin shell, inside: $E = 0$. Solid ball, inside: $E = \dfrac{Qr}{4\pi\varepsilon_0 R^3}$.
• Long wire: $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$. Large sheet: $E = \dfrac{\sigma}{2\varepsilon_0}$, at any distance.
• Two opposite sheets: $\dfrac{\sigma}{\varepsilon_0}$ between them and zero outside.
• Conductor: $E = 0$ inside, the charge is on the surface, and just outside $E = \dfrac{\sigma}{\varepsilon_0}$. A cavity is shielded.

Check yourself

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

1. Why is the field zero everywhere inside a thin charged spherical shell?

2. A large charged sheet is $2$ m from a point $P$. A second point $Q$ is $6$ m from the sheet on the same side. Compare the fields at the two points.

3. A long charged wire makes a field of $90$ N/C at a point. What is the field at a point three times as far from the wire?

4. The field just outside the surface of a charged conductor is $2700$ N/C. What is the charge on each square metre of the surface?

5. A hollow metal ball is given a charge. There is nothing inside the hollow. Where does the charge sit, and what is the field in the hollow?