A charge in a hollow
Take a thick neutral metal shell with a hollow inside it, a cavity. Place a charge \(q\) in the cavity, held on an insulating thread so that it does not touch the metal. Now draw a Gaussian surface inside the metal of the shell, all the way round the cavity. The field is zero at every point of it, so the charge it encloses is zero.
That surface encloses \(q\) and the charge on the cavity wall. So the wall must carry \(-q\). The shell is neutral, and its interior metal holds no charge, so the outer surface must carry \(+q\). If the shell had its own charge \(Q\), the outer surface would carry \(Q + q\). The inner wall's charge is always \(-q\), whatever \(Q\) is.
A point charge \(q = 2\,\mu\text{C}\) sits at the centre of a neutral thick metal shell of inner radius \(5\,\text{cm}\) and outer radius \(10\,\text{cm}\). Find the charge on each surface, and the field at \(2\,\text{cm}\), in the metal and at \(20\,\text{cm}\). Then find the potentials of the shell and of a point \(2.5\,\text{cm}\) from the centre. Take the \(k\) route.
The wall carries \(-2\,\mu\text{C}\) and the outer surface \(+2\,\mu\text{C}\). With \(q\) at the centre, both spread evenly, so each acts as a thin shell.
At \(2\,\text{cm}\) only \(q\) is inside a sphere through the point: \(E\) \(= kq/r^2\). In the metal, \(E = 0\). At \(20\,\text{cm}\) the enclosed charge is \(q - q + q = q\), so again \(E = kq/r^2\).
\(E\) \(= 1.8 \times 10^4/(0.02)^2\) \(= 4.5 \times 10^7\,\text{N/C}\) at \(2\,\text{cm}\), and \(1.8 \times 10^4/0.04\) \(= 4.5 \times 10^5\,\text{N/C}\) at \(20\,\text{cm}\). The shell is at \(kq/b\) \(= 1.8 \times 10^4/0.1\) \(= 1.8 \times 10^5\,\text{V}\).
At \(2.5\,\text{cm}\) all three charges add: \(V\) \(= kq/r - kq/a + kq/b\) \(= 1.8 \times 10^4(40 - 20 + 10)\) \(= 5.4 \times 10^5\,\text{V}\).
A metal body with a cavity carries a total charge of \(+7\,\text{nC}\). A charge of \(-5\,\text{nC}\) hangs inside the cavity, not touching the metal. How much charge sits on the body's outer surface, in nC?
The outside does not see where the charge is
Move the slider of the figure above. The negative charge on the wall crowds towards \(q\), wherever it goes. The charge \(q\) and the wall's charge together make no field at all beyond the cavity; they screen each other. So the outer surface feels nothing of where \(q\) sits. Its \(+2\,\mu\text{C}\) stays spread evenly, and the field outside stays \(kq/r^2\), centred on the shell.
Now choose the chip that earths the shell. The outer \(+2\,\mu\text{C}\) flows away to the ground, and the field outside falls to zero. The \(-2\,\mu\text{C}\) on the wall stays, held by the attraction of \(q\). Inside the cavity nothing changes at all. The earthed shell now hides the charge completely from the outside world.
A charge sits at the centre of the cavity of an isolated neutral metal shell. It is moved closer to one side of the cavity, without touching the wall. Which statements are true?
An empty cavity has no field
Take the charge out of the cavity. A Gaussian surface in the metal still encloses zero, so the wall's total charge is zero. Could the wall carry a plus patch and an equal minus patch? Then a field line would run across the cavity from one to the other. The potential falls along a field line, yet both ends are on the one equipotential conductor. So there are no patches, and the empty cavity has no field.
This holds whatever charges or fields are outside, whatever the shape of the cavity, and whatever the thickness of the metal. Outside charges only rearrange the charge on the outer surface. The cavity is sealed off from the electric world outside. This is electrostatic shielding, and a closed metal box used for it is a Faraday cage.
A thicker or better-conducting shell does not shield a steady field any better. Any closed metal skin, however thin, gives a field-free cavity once its charges settle. Thickness matters only for fields that change very fast, such as radio waves. Even then, what matters is how far a fast field reaches into the metal before it dies away.
The field inside a cavity of a conductor that holds no charge is zero, whatever the charges and fields outside the conductor.
Now the lift and the car of Lesson 1. A car body is a closed metal box. If lightning strikes it, the charge and the current stay on the outer skin and pass round the outside to the ground. Inside, the field stays nearly zero. A Delhi Metro coach is a metal box in the same way.
A phone call is carried by radio waves, fields that swing back and forth about a billion times a second. The free electrons of a metal wall follow these fields too and cancel them inside. A steel lift is not fully sealed, so the signal is weakened rather than cut off. That is why a call often drops but sometimes survives.
A radio is placed inside a closed metal box, and it stops receiving. A charged rod is then brought near the box. What is the steady electric field inside the box?
A hollow conductor in a uniform field
Place a hollow metal sphere in a uniform field pointing to the right. Free electrons drift left, against the field. They gather on the left face, the face the field points away from. The right face is left positive. In Fig 3.1 the field lines arriving from the left end on the negative charges. They start again from the positive charges on the right.
Outside, the lines bend so that they meet the metal at right angles, as Lesson 1 requires. Inside the metal and in the cavity, the field of the induced charges cancels the applied field completely. So the readout falls from \(1000\,\text{N/C}\) to zero. The cavity is shielded, although the conductor has no net charge.
For a metal sphere, solid or hollow, the induced charge per square metre is \(\sigma = 3\varepsilon_0E_0\cos\theta\). Here \(\theta\) is measured from the direction of the field. Just outside the two poles the field is \(3E_0\). The sphere gathers the field lines from a region three times its cross-section. That is why the lines in the figure bend in from so far away.
A hollow metal box stands in a uniform field that points to the right. Which statement is correct?
In 1836 Michael Faraday had a cube built, twelve feet on each side. Its wooden frame was covered with a net of copper wire, paper and tin foil. He charged the cube so strongly that sparks flew from its outside. Then he went inside with his most sensitive instruments. They showed no sign of any electric effect. The metal-covered box is still called a Faraday cage.
- Putting the induced charge of a cavity on the outer surface only.Draw a Gaussian surface in the metal. It encloses zero, so the cavity wall carries exactly \(-q\).
- Expecting the outside field to shift when the charge in the cavity moves.The charge and the wall screen each other. The outer charge stays even, and the outside field is unchanged.
- Believing a thicker metal wall shields a steady field better.Any closed metal skin gives a field-free cavity. Thickness matters only for fast-changing fields.