Concentric Shells and Earthing

Shells inside shells

Two thin metal shells share a centre. Shell A has radius \(a = 10\,\text{cm}\) and charge \(+2\,\mu\text{C}\). Shell B, outside it, has radius \(b = 20\,\text{cm}\) and charge \(+4\,\mu\text{C}\). These are concentric shells. A has nothing inside it, so all of its charge sits on its outer surface. B has two surfaces, inner and outer, and its charge divides between them.

Draw a Gaussian sphere inside the metal of B, the dashed sphere of Fig 4.1. The field in the metal is zero, so it encloses no net charge. It encloses A's \(+2\,\mu\text{C}\) and B's inner surface, so the inner surface carries \(-2\,\mu\text{C}\). B's total is \(+4\,\mu\text{C}\), so its outer surface carries \(4 - (-2)\) \(= +6\,\mu\text{C}\).

The two rules work for any number of shells. The inner surface of a shell carries minus the total charge inside it. Its outer surface carries whatever is left of the shell's own charge. The outermost surface of all therefore carries the total charge of the whole set. Outside everything, the field is that of the total charge placed at the centre.

Adding up potentials

Potential is a plain sum over charges, with no directions. So the potential at any point is the sum of the potentials of all the shells. From Lesson 2, a shell of radius \(R\) and charge \(Q\) adds \(kQ/r\) at a point outside it. At a point inside it, it adds its surface value \(kQ/R\), whatever the point.

Only each shell's total charge enters, because both of B's surfaces are at the same distance \(b\) from the centre. Shell A is inside B, so B adds \(kQ_B/b\) at A. Shell B is outside A, so A adds \(kQ_A/b\) at B.

For the example, \(V_A - V_B\) \(= 9 \times 10^9 \times 2 \times 10^{-6}(10 - 5)\) \(= 9 \times 10^4\,\text{V}\), as found. Give B any charge at all, and this difference stays the same. Only the charge on the inner shell sets the difference. This idea returns in the chapter Capacitance and Dielectrics as the spherical capacitor.

What earthing fixes

The Earth is a conductor so large that a few microcoulombs change its potential by almost nothing. Adding \(1\,\mu\text{C}\) to it shifts its potential by about \(kQ/R_E\) \(\approx 10^{-3}\,\text{V}\). So the Earth's potential is taken as zero, the reference for every other potential. Joining a conductor to the Earth by a wire is earthing. The two become one conductor, so the conductor comes to zero potential.

Earth shell B of the example. Its potential becomes zero: \(k(Q_A + Q_B)/b = 0\), so \(Q_B = -Q_A = -2\,\mu\text{C}\). The \(+6\,\mu\text{C}\) of B's outer surface flows to the ground. The \(-2\,\mu\text{C}\) on its inner surface stays, held by A. Then \(V_A\) \(= kQ_A(1/a - 1/b)\) \(= 9 \times 10^4\,\text{V}\): the difference was fixed, so A sits \(9 \times 10^4\,\text{V}\) above zero.

Now earth A instead, and leave B with \(+4\,\mu\text{C}\). A must reach zero: \(kQ_A/a + kQ_B/b = 0\), so \(Q_A\) \(= -Q_B\,a/b\) \(= -4 \times 0.1/0.2\) \(= -2\,\mu\text{C}\). B's charge raised A's potential, so negative charge flowed up from the ground to cancel it. A was not left neutral. Choose the chips of Fig 4.1 to compare the four cases.

That is the earth pin of the plug. It fixes the metal body of an iron or a refrigerator at the potential of the ground. If a live wire touches the body inside, current rushes to earth, and the fuse or breaker cuts the supply. A person who touches the body is then at nearly the potential of the floor, so almost no current flows through them.

Joining conductors

Join shells A and B by a thin wire. They become one conductor, and all the charge of a conductor sits on its outer surface. So A ends with no charge, and B's outer surface carries all \(6\,\mu\text{C}\). Both are at \(9 \times 10^9 \times 6 \times 10^{-6}/0.2\) \(= 2.7 \times 10^5\,\text{V}\). This is the last chip of Fig 4.1.

With radii \(10\,\text{cm}\) and \(20\,\text{cm}\), the figure above shows the numbers. Before the switch closes, the small sphere is at \(kQ/R\) \(= 5.4 \times 10^5\,\text{V}\) and the large one at zero. Charge flows until both are at \(1.8 \times 10^5\,\text{V}\), with \(2\,\mu\text{C}\) and \(4\,\mu\text{C}\). Some energy is lost as heat in the wire; the chapter Capacitance and Dielectrics counts it.