Conductors and dielectrics in a field

Why a car is safe in a thunderstorm

Lightning hits a car. The people inside are not hurt, because the metal body did the job. How can a thin metal shell protect everything inside it? The answer is in how charges behave in a conductor. We shall also meet the opposite kind of material, the dielectric, which is the key to capacitors.

Conductors and insulators

A conductor has charges that are free to move. A metal has free electrons. An insulator has almost none, because every charge is held in its own atom or molecule. Glass, plastic, dry wood and air are insulators.

A conductor at rest

Place a conductor in an electric field, or give it some charge. The free charges move for a very short time. Then they stop. We say the conductor is in electrostatic equilibrium. Four facts hold.

1. The field inside is zero. Suppose there were a field. It would push the free charges, and they would keep moving. They stop only when the field is gone.
2. There is no charge inside. Any extra charge sits on the surface. Take any small closed surface inside the metal. The field is zero on it, so by Gauss's law the charge inside it is zero. Charges of the same sign push each other as far apart as they can go, which is the outer surface.
3. The whole conductor has one potential. Between two points inside, $V = \int E \, dl$ and $E = 0$, so there is no difference. The surface is an equipotential too.
4. The field just outside is at right angles to the surface, and its size is $\sigma/\varepsilon_0$. If the field had a part along the surface, charges on the surface would slide along. So only the part at right angles is left.

The last fact needs a short proof. Draw a small flat box that sticks half out of the surface, with a face area $\Delta A$. Only the outer face has a field through it, and the field inside is zero. Gauss's law gives $E\,\Delta A = \sigma\,\Delta A/\varepsilon_0$, so

Here $\sigma$ is the charge on each unit of area of the surface, and $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)$.

The left picture shows a metal body in a uniform field. The minus signs on the left and the plus signs on the right cancel the field in the body. The lines end and start on the surface, always at right angles.

A worked example. A metal ball of radius $15$ cm carries $6.0\ \mu\text{C}$. Find the field just outside it.

1. The charge on each unit of area. $\sigma = \dfrac{Q}{4\pi R^2} = \dfrac{6.0 \times 10^{-6}\ \text{C}}{4\pi \times (0.15\ \text{m})^2} = 2.12 \times 10^{-5}\ \text{C/m}^2$.
2. The field. $E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{2.12 \times 10^{-5}\ \text{C/m}^2}{8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 2.40 \times 10^{6}\ \text{V/m}$.
3. A check. The field outside a sphere is $\dfrac{kQ}{R^2} = \dfrac{9.0 \times 10^{9} \times 6.0 \times 10^{-6}}{(0.15)^2} = 2.4 \times 10^{6}\ \text{V/m}$. It fits.

Air breaks down and sparks at about $3 \times 10^{6}$ V/m. So this ball is close to sparking.

A hollow conductor shields what is inside

Take a metal box with a hollow inside. The field in the metal is zero. So the field in the cavity does not depend on charges or fields outside. Move a charge outside, or switch on a strong field, and the cavity stays at zero field. This is electrostatic shielding.

The same fact explains the car, and why a sensitive instrument is put in a metal case.

Now put a charge $+q$ inside the cavity. A charge $-q$ gathers on the inner wall, because the field lines of $+q$ must end there. The wall was neutral, so a charge $+q$ appears on the outer surface. The right-hand picture shows it.

A shell protects its inside from the outside. It does not hide a charge that sits inside. The field of that charge still shows outside, unless the shell is earthed.

Sharp points and the joined spheres

Join two metal spheres of different radius by a long thin wire, so that they are one conductor. Their potential is the same. A sphere of radius $R$ with charge $q$ has the potential $\dfrac{kq}{R}$, so

The bigger sphere keeps more charge. But look at the charge on each unit of area, $\sigma = \dfrac{q}{4\pi R^2}$:

The smaller sphere has the larger $\sigma$, and so the larger field $E = \sigma/\varepsilon_0$ at its surface. A sharp point is like a very small sphere. The charge crowds there, and the field is the strongest. A pointed rod on a roof, a lightning conductor, makes the air at its tip break down first, and leads the charge safely to the earth.

A worked example. Two spheres of radius $6.0$ cm and $18$ cm are joined by a long wire and carry $8.0$ nC together. Find the charge on each, and the potential.

1. The charges. $q_1 : q_2 = R_1 : R_2 = 6.0\ \text{cm} : 18\ \text{cm} = 1 : 3$. So $q_1 = \dfrac{1}{4} \times 8.0\ \text{nC} = 2.0\ \text{nC}$ and $q_2 = 6.0\ \text{nC}$.
2. The potential. $V = \dfrac{kq_1}{R_1}$, and $kq_1 = 9.0 \times 10^{9} \times 2.0 \times 10^{-9} = 18\ \text{N m}^2/\text{C}$. So $V = \dfrac{18\ \text{N m}^2/\text{C}}{0.060\ \text{m}} = 300\ \text{V}$.
3. The charge on each unit of area. The small sphere has $\dfrac{\sigma_1}{\sigma_2} = \dfrac{18\ \text{cm}}{6.0\ \text{cm}} = 3.0$ times as much.

Make the two radii equal. Is the charge shared equally? Then make sphere 2 very large. Where does almost all the charge go?

Dielectrics

An insulator also reacts to a field, but in a different way. Its charges cannot leave their molecules. A material used for this is called a dielectric. There are two kinds of molecule.

• Non-polar molecules, such as nitrogen or methane, have no dipole moment without a field. A field pulls the positive and the negative charges of each molecule a little apart, so each gets an induced dipole moment along the field.
• Polar molecules, such as water, always have a dipole moment, but they point every way at random. A field turns them so that on average they line up with it. Heat motion stops them from lining up completely.

Either way, the slab is polarised. Look at the picture.

Inside the slab the plus ends and the minus ends of neighbours meet and cancel. At the two faces nothing cancels. A layer of bound charge appears: negative on the face where the field enters, positive on the face where it leaves. The word bound means that these charges cannot leave the slab, unlike the free charge on a metal plate.

The bound charges make a field of their own, $E_p$, inside the slab. It points against the applied field $E_0$. So the field in the slab is smaller:

The ratio of the two fields is a number for each material, the dielectric constant $K$:

A dielectric never cancels the field completely. It only cuts it by the factor $K$, which is greater than $1$. For a vacuum $K = 1$, for air $K = 1.0006$, for mineral oil about $2$, for glass between $5$ and $10$, and for water about $80$. For a metal the field inside is zero, which is what $K = \infty$ would give. So a conductor acts like a dielectric of infinite constant.

How much bound charge is there? Let the free charge on each unit of area of a plate be $\sigma$, so $E_0 = \sigma/\varepsilon_0$. The bound charge on each face is $\sigma_b$, and it cancels part of it, $E = (\sigma - \sigma_b)/\varepsilon_0$. Put $E = E_0/K$:

A worked example. A slab of dielectric constant $5.0$ is placed in a field of $6.0 \times 10^{4}$ V/m at right angles to its faces. Find the field inside, and the charge on each unit of area of a face.

1. The field inside. $E = \dfrac{E_0}{K} = \dfrac{6.0 \times 10^{4}\ \text{V/m}}{5.0} = 1.2 \times 10^{4}\ \text{V/m}$.
2. The field of the faces. $E_p = E_0 - E = 4.8 \times 10^{4}\ \text{V/m}$.
3. The bound charge. The faces make $E_p = \sigma_b/\varepsilon_0$, so $\sigma_b = \varepsilon_0 E_p = 8.85 \times 10^{-12} \times 4.8 \times 10^{4} = 4.2 \times 10^{-7}\ \text{C/m}^2$.

Push $K$ up to $20$. Does the field inside ever reach zero? Now think of a metal.

Practise with new numbers

Try it in Python

The program joins two spheres by a wire. The first sphere always has a radius of $6.0$ cm. It tries four radii for the second sphere, and finds the charges, the potential and the ratio of the charge on each unit of area.

Common mistakes

• Saying there is no field near a conductor. The field is zero inside the metal. Just outside it is $\sigma/\varepsilon_0$.
• Putting the extra charge inside the metal. It sits on the surface. The inside of a solid conductor has no charge.
• Drawing field lines that meet the surface at a slant. In equilibrium they meet it at right angles.
• Thinking the bigger sphere has the larger charge on each unit of area. The bigger sphere holds more charge in all, but the smaller sphere holds more on each square metre.
• Thinking a dielectric cancels the field. It cuts it by the factor $K$. Only a conductor makes it zero.
• Using cm for a radius in $4\pi R^2$. Change to metres first.

Remember these

• In a conductor at rest: $E = 0$ inside, charge only on the surface, one potential, and the field outside at right angles to the surface.
• The field just outside: $E = \dfrac{\sigma}{\varepsilon_0}$.
• A closed hollow conductor shields its cavity. A charge $+q$ inside draws $-q$ on the inner wall and $+q$ on the outer wall.
• Joined spheres share charge in the ratio of the radii, $q_1/q_2 = R_1/R_2$. The charge on each unit of area goes as $1/R$, so it is largest at sharp points.
• A dielectric polarises, and the bound charge cuts the field inside to $E = E_0/K$. The bound charge is $\sigma_b = \sigma(1 - 1/K)$.

Check yourself

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

1. Why must the field inside a metal be zero when no current flows?

2. A metal can that is hollow and neutral holds a ball with a charge of $+12$ nC without touching it. What charge is on the inner wall, and what charge is on the outer wall?

3. The field just outside the flat surface of a charged conductor is $4.4 \times 10^{4}$ V/m. What is the charge on each square metre of the surface?

4. A slab placed at right angles to a field of $9.0 \times 10^{4}$ V/m cuts the field inside it to $1.5 \times 10^{4}$ V/m. What is its dielectric constant?