Potential of Bodies and Energy of Systems

A body is many point charges

Cut a charged body into small pieces of charge \(dq\). Each piece is nearly a point charge, and potentials add as numbers. So the potential of the body at a point P is \(V = \int k\,dq/r\), where \(r\) is the distance of each piece from P. No components and no directions enter. That makes potentials of bodies much easier than their fields.

A ring and an arc

Every piece of a ring of radius \(R\) is the same distance \(R\) from its centre. So \(r\) comes out of the integral, and \(V = kQ/R\), whatever the charge on the ring. A point on the axis at a distance \(x\) from the centre is \(\sqrt{R^2 + x^2}\) from every piece, so \(V = kQ/\sqrt{R^2 + x^2}\) there.

The same argument works for any part of a ring. A wire bent into a semicircle of radius \(10\,\text{cm}\) and carrying \(2\,\mu\text{C}\) gives \(kQ/R = 1.8 \times 10^5\,\text{V}\) at its centre. Bent into a full circle, it gives the same. Only the charge and the radius matter, not how the charge is spread along the arc.

At the centre the field of each piece is cancelled by the piece opposite, so \(E = 0\). The potentials of those same pieces add, because each is a positive number. So \(V = kQ/R\) is largest just where \(E = 0\). At \(x = 40\,\text{cm}\) every piece is \(50\,\text{cm}\) away, and \(V = 9 \times 10^9 \times 5 \times 10^{-6}/0.5 = 9 \times 10^4\,\text{V}\).

A charged shell

A thin spherical shell of radius \(R\) carries a charge \(Q\). Every piece of it is \(R\) from the centre, so the potential at the centre is \(kQ/R\). The shell theorem of the Gravitation chapter holds for any inverse-square force. So the field inside the shell is zero, and outside the shell acts like a point charge \(Q\) at its centre.

A zero field inside means that no work is done in moving a charge about inside. So the potential is the same everywhere inside as at the centre, \(kQ/R\), which is also the value on the surface. Outside it falls as \(kQ/r\). For \(2\,\mu\text{C}\) on a shell of radius \(10\,\text{cm}\): \(1.8 \times 10^5\,\text{V}\) on and inside it, and \(9 \times 10^4\,\text{V}\) at \(20\,\text{cm}\).

Fig 3.1 shows the link between the two graphs. Where \(E = 0\) the potential is flat; where \(E\) is large the potential falls steeply. The next lesson turns this into a rule: the field is minus the slope of the potential.

A rod, by an integral

For a rod the pieces are at different distances, and the integral must be done.

A rod \(30\,\text{cm}\) long with \(\lambda = 2\,\mu\text{C/m}\), seen from a point \(10\,\text{cm}\) beyond one end on its line, gives \(V = 9 \times 10^9 \times 2 \times 10^{-6} \times \ln 4\) \(= 1.8 \times 10^4 \times 1.39\) \(\approx 2.5 \times 10^4\,\text{V}\).

A disc, from rings

A disc of radius \(R\) with surface charge density \(\sigma\) is a set of rings. A ring of radius \(s\) and width \(ds\) holds \(dq = \sigma\,2\pi s\,ds\), and on the axis it is \(\sqrt{s^2 + x^2}\) from P. Adding the rings from \(s = 0\) to \(s = R\) gives the result below. Route: \(\varepsilon_0\), as for every sheet.

\[ V = \dfrac{\sigma}{2\varepsilon_0}\left(\sqrt{x^2 + R^2} - x\right) \]

For \(\sigma = 1.77 \times 10^{-8}\,\text{C/m}^2\), the factor is \(\sigma/2\varepsilon_0\) \(= 1.77 \times 10^{-8}/(2 \times 8.85 \times 10^{-12})\) \(= 1000\,\text{V/m}\). A disc of radius \(30\,\text{cm}\) then has \(V = 1000 \times 0.3 = 300\,\text{V}\) at its centre. At \(x = 40\,\text{cm}\) it has \(V = 1000 \times (0.5 - 0.4) = 100\,\text{V}\).

Building a system charge by charge

How much work does it take to assemble several charges? Bring them in from infinity one at a time. The first costs nothing, because no other charge is there yet. The second works against the first. The third works against the first two, and so on. Each new charge pays once for its pair with every charge already present.

In Fig 3.2 every side of the triangle is a pair worth \(kq^2/a = 9 \times 10^9 \times 4 \times 10^{-12}/0.2\) \(= 0.18\,\text{J}\). Three pairs give \(0.54\,\text{J}\). With the third charge reversed, two pairs become \(-0.18\,\text{J}\) and the total is \(0.18 - 0.36 = -0.18\,\text{J}\). The order of assembly does not matter: every order counts each pair exactly once.

To take one charge away, undo only the pairs it belongs to. Put \(2\,\mu\text{C}\) at A, \(-1\,\mu\text{C}\) at B and \(2\,\mu\text{C}\) at C, on a line with \(10\,\text{cm}\) gaps. The pairs AB and BC are \(-0.18\,\text{J}\) each, and AC is \(+0.18\,\text{J}\), so \(U = -0.18\,\text{J}\). Removing C undoes AC and BC, which add to zero, so taking C to infinity costs no work.

The energy of the charge on a shell

A single charged body is also a system of charges: all the small pieces on it repel one another. The energy needed to gather its charge from infinity is its self-energy. For a thin shell, bring the charge in small amounts. When the shell already holds \(q\), its potential is \(kq/R\), and the next piece \(dq\) costs \(kq\,dq/R\).

For \(2\,\mu\text{C}\) on a shell of radius \(10\,\text{cm}\), \(U = 9 \times 10^9 \times 4 \times 10^{-12}/0.2 = 0.18\,\text{J}\). The energy grows as \(Q^2\). Doubling the charge makes it \(0.72\,\text{J}\), so the second \(2\,\mu\text{C}\) costs \(0.54\,\text{J}\), three times the first.

A solid sphere of the same charge and radius has \(3kQ^2/5R = 0.216\,\text{J}\), more than the shell. Inside a solid sphere much of the charge sits closer together than on a shell, so gathering it costs more work.

Put a point charge of \(1\,\mu\text{C}\) at the centre of the \(2\,\mu\text{C}\) shell. Every piece of the shell is \(R\) away, so the interaction energy is \(kQq/R = 0.18\,\text{J}\). The total electrostatic energy is the shell's self-energy plus this, \(0.36\,\text{J}\). A point charge has no finite self-energy, so it is left out; it never changes in any process anyway.