A charged metal sphere
Give a metal sphere of radius \(R\) a charge \(Q\). The charge goes to the surface, and on a lone sphere no point of the surface is special. So the charge spreads evenly, and the sphere is a thin charged shell. The results of the chapter Electric Flux and Gauss's Law then apply at once. There is no field inside, and outside is the field of a point charge \(Q\) at the centre.
The potential outside is that of a point charge too, \(kQ/r\). At the surface it is \(kQ/R\). Inside, the field is zero, so the potential does not change as you move in. It stays \(kQ/R\) all the way to the centre. The potential has no jump at the surface; only the field jumps there.
A solid metal sphere of radius \(10\,\text{cm}\) is given \(2\,\mu\text{C}\). Find the field at its centre, at \(5\,\text{cm}\), just outside its surface and at \(20\,\text{cm}\). Then find the potential at its surface and at its centre. Take the \(k\) route.
At the centre and at \(5\,\text{cm}\) the point is inside the metal, so \(E = 0\).
Just outside, \(E\) \(= kQ/R^2\) \(= 9 \times 10^9 \times 2 \times 10^{-6}/0.01\) \(= 1.8 \times 10^6\,\text{N/C}\). At \(20\,\text{cm}\), \(E\) \(= 1.8 \times 10^4/0.04\) \(= 4.5 \times 10^5\,\text{N/C}\).
At the surface, \(V\) \(= kQ/R\) \(= 1.8 \times 10^4/0.1\) \(= 1.8 \times 10^5\,\text{V}\). The centre is at the same potential, since no field acts inside.
Values used: \(k\) \(= 9 \times 10^9\,\text{N m}^2/\text{C}^2\).
The \(\varepsilon_0\) route gives the same field. The charge per square metre is \(\sigma\) \(= Q/4\pi R^2\) \(\approx 1.59 \times 10^{-5}\,\text{C/m}^2\). Just outside, \(E\) \(= \sigma/\varepsilon_0\) \(\approx 1.8 \times 10^6\,\text{N/C}\), as in Lesson 1. The two routes differ by half a percent, as the rounded \(k\) and \(\varepsilon_0\) do.
The metal sphere of the example carries \(2\,\mu\text{C}\) on a radius of \(10\,\text{cm}\). What is the potential \(30\,\text{cm}\) from its centre, in V? Take \(k\) \(= 9 \times 10^9\,\text{N m}^2/\text{C}^2\).
At the centre of the charged metal sphere of the example, which pair of values is right?
Where the surface curves sharply
A lone sphere spreads its charge evenly. A body of any other shape does not. To see why, join a small sphere of radius \(r_1\) and a large one of radius \(r_2\) by a long thin wire. Put them far apart, so that each acts as a lone sphere. The wire makes them one conductor, and one conductor has one potential.
\[ \dfrac{kQ_1}{r_1} = \dfrac{kQ_2}{r_2} \]
Why. the two spheres are at one potential, and each is far enough from the other to count as a lone sphere.
\[ \dfrac{Q_1}{Q_2} = \dfrac{r_1}{r_2} \]
Why. cancel \(k\) and rearrange. The charges go as the radii.
\[ \dfrac{\sigma_1}{\sigma_2} = \dfrac{Q_1/4\pi r_1^2}{Q_2/4\pi r_2^2} = \dfrac{r_1}{r_2}\cdot\dfrac{r_2^2}{r_1^2} \]
Why. the charge per square metre is the charge divided by the surface area \(4\pi r^2\).
\[ \dfrac{\sigma_1}{\sigma_2} = \dfrac{r_2}{r_1} \]
Read it. the ratio simplifies. The smaller sphere carries less charge but on far less area, so its surface is more densely charged.
So \(\sigma \propto 1/r\): the smaller sphere has the denser charge. Just outside, \(E = \sigma/\varepsilon_0\), so the field is stronger at the smaller sphere as well. A pear-shaped conductor behaves like a large sphere at its blunt end joined to a small sphere at its sharp end. Its charge crowds towards the sharp end.
Charge is placed evenly over a pear-shaped metal body and then left to settle. Where does it end up most densely packed?
Play the figure above. The ends of the drawn pear have radii in the ratio \(1 : 0.28\). Treating them as two joined spheres, the charge density at the sharp end becomes about \(1/0.28 \approx 3.6\) times that at the blunt end. The field arrows grow at the tip in the same ratio. The count of signs does not change, because the charge is only moving round the surface.
A sharp point is a very small radius, so the field just outside a point can be enormous. When it passes about \(3 \times 10^6\,\text{V/m}\), stray electrons in the air knock more electrons out of molecules, and the air there conducts. The point then leaks its charge into the air with a faint glow and a hiss. This leak is a corona discharge; Lesson 5 puts numbers on it.
A charged metal body has a flat face, a gently rounded edge and a sharp corner. Where is the field just outside it strongest?
In the 1740s Benjamin Franklin noticed that a sharp needle drew the charge silently off a charged metal ball. A blunt rod drew it only with a spark, from close by. In 1750 he proposed that pointed iron rods, run down into the ground, could protect buildings from lightning. He hoped the points would drain a cloud quietly. They cannot do that, but they give a strike a safe path to the ground, and his rod is still in use.
Tall buildings, towers and temples in India carry a lightning conductor. It is a pointed metal rod above the highest point. A thick copper strip joins it to a metal plate buried in damp earth. Under a charged thundercloud, strong fields gather at the point, and it leaks charge steadily into the air. If lightning still strikes, the rod gives the current a path of low resistance to the ground, outside the walls.
Two spheres at one potential
Now put numbers on two spheres, of radii \(10\,\text{cm}\) and \(20\,\text{cm}\), each raised to \(9000\,\text{V}\). The potential of a sphere is \(kQ/R\), so \(Q = VR/k\): at one potential, charge is in proportion to the radius. It makes no difference whether the spheres are joined by a wire or charged separately to the same potential.
Two isolated metal spheres, of radii \(10\,\text{cm}\) and \(20\,\text{cm}\), are each raised to \(9000\,\text{V}\). Find the charge on each and the ratio of their surface charge densities. Take the \(k\) route.
Each sphere is a lone sphere, so \(V = kQ/R\), which gives \(Q = VR/k\).
\(Q_1\) \(= 9000 \times 0.1/(9 \times 10^9)\) and \(Q_2\) \(= 9000 \times 0.2/(9 \times 10^9)\).
\(Q_1\) \(= 10^{-7}\,\text{C}\) \(= 100\,\text{nC}\), and \(Q_2 = 200\,\text{nC}\). With \(\sigma \propto 1/R\), \(\sigma_1/\sigma_2\) \(= 20/10\) \(= 2\).
The spheres hold \(100\,\text{nC}\) and \(200\,\text{nC}\). The smaller one has twice the charge density, so twice the field just outside it.
Two metal spheres of radii \(5\,\text{cm}\) and \(15\,\text{cm}\), far apart, are joined by a thin wire and share \(8\,\mu\text{C}\). How much charge ends on the smaller sphere, in μC?
- Setting the potential inside a charged metal sphere to zero because the field there is zero.Zero field means a constant potential. Inside, \(V\) keeps its surface value \(kQ/R\).
- Assuming that one potential means one charge density.Joined bodies share one potential, but \(\sigma \propto 1/r\). The charge is densest where the surface curves most.
- Sharing charge equally between two joined spheres.Equal potential gives \(Q_1/r_1 = Q_2/r_2\), so the larger sphere takes the larger share.