Step into a steel lift in a Mumbai office tower with a phone call running, and the call often drops before the doors open again. Out on the highway, a thunderstorm breaks over a car. People are told to stay inside the closed car rather than run for a tree. The two pieces of advice look unrelated. Both rest on what metal does to an electric field.
Charge that is free to move
In a metal, each atom gives up one or two of its outer electrons. These electrons are not tied to any one atom and wander through the whole piece. They are the free electrons, and copper has about \(8.5 \times 10^{28}\) of them in every cubic metre. The positive ions stay fixed in their places. A material with charge free to move through it is a conductor.
A metal slab in a field
Place a metal slab in a uniform field \(\vec E_0\) pointing to the right. Each free electron feels a force \(-e\vec E_0\), to the left, and drifts that way. Electrons gather on the left face, which becomes negative. The right face, which has lost electrons, is left positive. This separated charge is the induced charge; the slab as a whole stays neutral.
The two layers of induced charge act like two charged sheets. Between them, inside the metal, their field \(\vec E_\text{ind}\) points from the positive face to the negative face, against \(\vec E_0\). As long as \(E_\text{ind}\) is smaller than \(E_0\), a net field remains and the electrons keep drifting. So the drift stops only when \(E_\text{ind} = E_0\) and the net field inside is zero.
In Fig 1.1 the field is \(1000\,\text{N/C}\). Two sheets with \(\pm\sigma\) give \(\sigma/\varepsilon_0\) between them, so the faces need \(\sigma\) \(= \varepsilon_0 E_0\) \(= 8.85 \times 10^{-9}\,\text{C/m}^2\). That is about one extra electron for every few hundred million atoms of the surface. A tiny shift of charge is enough to cancel the field completely.
When the charges of a conductor are at rest, the electric field at every point inside its material is zero. This state is electrostatic equilibrium.
The rule holds only when nothing moves, which is what electrostatics means. A wire connected to a cell carries a steady current, and there is a field inside it that drives the current. That case belongs to the chapter Electric Current and Resistance. In this chapter every charge has come to rest.
Why must the electric field inside a metal be zero once its charges are at rest?
Where the extra charge goes
Give a metal body some extra charge and let it settle. Draw a Gaussian surface inside the metal, as close to the outer surface as you like; it is S in Fig 1.2. The field is zero at every point of S, so the flux through S is zero. By Gauss's law, S encloses no net charge. The extra charge cannot be inside S, so it must sit on the surface.
The inside of the metal is not empty of charge. Every small piece still holds ions and electrons in equal numbers, so it is neutral. The extra charge lives in a layer only a few atoms thick at the surface. This is why a hollow metal ball and a solid one of the same size hold the same charge in the same way.
A solid copper ball is given a charge of \(+5\,\mu\text{C}\) and left alone. Which statements are true once it has settled?
One potential for the whole body
The potential difference between two points is the work per unit charge done by the field on the way. Inside a conductor the field is zero, so this work is zero along any path within the metal. Every point of the metal, including the surface, is therefore at the same potential. A conductor in electrostatic equilibrium is an equipotential body.
This gives a quick way to find the potential of a conductor: find it at the most convenient point. The centre of a sphere is often that point. Every charge on the sphere's surface is at the same distance \(R\) from the centre, whatever its pattern.
An uncharged metal sphere of radius \(10\,\text{cm}\) stands with its centre \(30\,\text{cm}\) from a point charge \(q = +2\,\mu\text{C}\). Find the potential of the sphere and the field of the induced charges at its centre. Take the \(k\) route.
The charge \(q\) draws electrons to the near side, leaving the far side positive. The induced charges add up to zero, because the sphere stays neutral.
At the centre, \(V\) \(= kq/r + k\,q_\text{ind}/R\). Every induced charge is at the distance \(R\), and they add up to zero, so the second term vanishes.
\(V\) \(= 9 \times 10^9 \times 2 \times 10^{-6}/0.3\) \(= 6 \times 10^4\,\text{V}\). This is the potential of every point of the sphere. The net field at the centre is zero, so the induced charges give \(kq/r^2\) \(= 2 \times 10^5\,\text{N/C}\) there, pointing towards \(q\).
The sphere is at \(6 \times 10^4\,\text{V}\), and its induced charges make \(2 \times 10^5\,\text{N/C}\) at its centre, pointing towards the charge.
An uncharged metal sphere has its centre \(45\,\text{cm}\) from a point charge of \(+3\,\mu\text{C}\). What is the potential of the sphere, in V? Take \(k\) \(= 9 \times 10^9\,\text{N m}^2/\text{C}^2\).
The field at the surface
Just outside a conductor, the field must point along the normal to the surface. Suppose it had a part along the surface. That part would push the surface charges along the surface, and they would move. At equilibrium nothing moves, so the part along the surface is zero. Equally, a field along an equipotential surface is impossible.
Now find the size of that field from the surface charge density \(\sigma\) at the spot. Take a small pillbox of face area \(A\), with one face just outside and one inside the metal, as in Fig 1.2.
\[ \Phi_\text{inner face} = 0,\quad \Phi_\text{side} = 0 \]
Why. the face inside the metal has no field on it. Along the short side, the field is normal to the surface and so runs parallel to the side.
\[ \Phi_\text{outer face} = EA \]
Why. the field just outside crosses the outer face along its normal and has one size over a small patch.
\[ EA = \dfrac{\sigma A}{\varepsilon_0} \quad\Rightarrow\quad E = \dfrac{\sigma}{\varepsilon_0} \]
Read it. Gauss's law, with the pillbox enclosing the charge \(\sigma A\) of the patch of surface it crosses.
For \(\sigma\) \(= 8.85 \times 10^{-9}\,\text{C/m}^2\), the field just outside is \(8.85 \times 10^{-9}/8.85 \times 10^{-12}\) \(= 1000\,\text{N/C}\). A thin charged sheet with the same \(\sigma\) gives only \(500\,\text{N/C}\). The sheet sends its flux out on both sides, \(\sigma/2\varepsilon_0\) each way. A conductor has no field inside, so all of its flux leaves through the outer side.
A thin metal plate given a charge \(Q\) shows that the two results agree. The charge splits equally between the plate's two faces, \(Q/2\) on each. Each face then has \(\sigma/2\), where \(\sigma\) is the plate's total charge per square metre. So the field outside is \((\sigma/2)/\varepsilon_0\) \(= \sigma/2\varepsilon_0\), exactly the thin sheet's field.
The field just outside a charged conductor is \(4.5 \times 10^5\,\text{N/C}\). What is the surface charge density there, in μC/m²? Take \(\varepsilon_0\) \(= 8.85 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)\).
Two large parallel metal plates carry total charges \(Q\) and \(2Q\). The field inside the metal of each plate is zero. What charges sit on the four faces, from left to right?
With this, the phone in the lift is half explained. No steady field can exist inside the metal of the lift's walls. Lesson 3 shows that no field can exist in the space the walls enclose either, and that settles both the lift and the car.
- Thinking that the metal blocks the outside field from entering.The outside field enters; the induced charges make an equal and opposite field, and the two cancel inside.
- Using \(\sigma/2\varepsilon_0\) just outside a conductor.A conductor sends all its flux outward, so \(E = \sigma/\varepsilon_0\). The value \(\sigma/2\varepsilon_0\) belongs to a thin sheet.
- Placing extra charge throughout the volume of a metal body.A Gaussian surface inside the metal has zero field, so it encloses no net charge. The extra charge is on the surface.