IPhO 1970 · Theory problem 3

Earthed metal ball inside a charged spherical shell

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Inside a thin-walled metal sphere with radius \(R=20\ \mathrm{cm}\) there is a metal ball with the radius \(r = 10\ \mathrm{cm}\) which has a common centre with the sphere. The ball is connected with a very long wire to the Earth via an opening in the sphere (Fig. 3). A charge \(Q = 10^{-8}\ \mathrm{C}\) is placed onto the outside sphere. Calculate the potential of this sphere, electrical capacity of the obtained system of conducting bodies and draw out an equivalent electric scheme.

Figure 1

Fig. 3

Official English text of the problem.

Answers

Problem 3 has no numbered subparts. In the requested order:

  • Potential of the outer sphere relative to Earth: \[ \boxed{V=\frac{Q(R-r)}{4\pi\varepsilon_0R^2}\approx225\ \mathrm{V}.} \]
  • Electrical capacity of the system: \[ \boxed{C=\frac{4\pi\varepsilon_0R^2}{R-r} \approx4.45\times10^{-11}\ \mathrm{F}=44.5\ \mathrm{pF}.} \]
  • Equivalent electric scheme: two capacitors connected in parallel between the outer sphere and Earth: \[ \boxed{C_{\mathrm{gap}}=\frac{4\pi\varepsilon_0rR}{R-r}\approx22.3\ \mathrm{pF}, \qquad C_{\mathrm{ext}}=4\pi\varepsilon_0R\approx22.3\ \mathrm{pF}.} \]

Solution

We neglect the disturbance caused by the small opening and the connecting wire, and treat Earth as a distant conductor at zero potential. The shell thickness is negligible, so both its surfaces have radius \(R\).

The outer shell is electrically isolated and retains total charge \(Q\). The inner ball, however, can exchange charge with Earth.

Potential of the outer sphere.

Let the final charge on the inner ball be \(q\). In electrostatic equilibrium, the electric field inside the metal of the shell vanishes. A Gaussian surface within that metal therefore encloses zero net charge. Consequently, the shell’s inner surface carries charge \(-q\), and its outer surface carries \[ Q_{\mathrm{outer}}=Q+q. \]

Write \(k=1/(4\pi\varepsilon_0)\). A uniformly charged spherical surface produces constant potential \(kQ/R\) everywhere inside it. Thus the potential of the inner ball is \[ V_{\mathrm{ball}}=k\left(\frac{q}{r}+\frac{Q}{R}\right). \] Because the ball is grounded, \[ V_{\mathrm{ball}}=0 \quad\Longrightarrow\quad q=-Q\frac{r}{R}. \] Numerically, \[ q=-5.0\times10^{-9}\ \mathrm{C}. \] The inner and outer surfaces of the shell therefore each carry \(+5.0\times10^{-9}\ \mathrm{C}\).

At the shell, the potential is \[ V=k\frac{Q+q}{R} =k\frac{Q}{R}\left(1-\frac{r}{R}\right) =\frac{Q(R-r)}{4\pi\varepsilon_0R^2}. \] Using \(k\simeq8.99\times10^9\ \mathrm{N\,m^2\,C^{-2}}\), \[ V=(8.99\times10^9)(10^{-8}) \frac{0.20-0.10}{(0.20)^2} \approx\boxed{225\ \mathrm{V}}. \]

Electrical capacity of the system.

The capacitance is the charge placed on the isolated shell divided by its potential relative to Earth: \[ C=\frac{Q}{V} =4\pi\varepsilon_0\frac{R^2}{R-r}. \] Hence \[ C=\frac{1}{8.99\times10^9} \frac{(0.20)^2}{0.20-0.10} \approx\boxed{4.45\times10^{-11}\ \mathrm{F}=44.5\ \mathrm{pF}}. \]

Equivalent electric scheme.

There are two contributions to the charge stored on the shell.

The field in the gap corresponds to a spherical capacitor between the shell and the grounded ball. Its potential difference is \[ V= k(-q)\left(\frac1r-\frac1R\right), \] so its capacitance is \[ C_{\mathrm{gap}}=\frac{-q}{V} =\frac{4\pi\varepsilon_0rR}{R-r}. \]

The exterior field corresponds to a capacitor between the shell’s outer surface and distant Earth. Since \[ V=k\frac{Q_{\mathrm{outer}}}{R}, \] this contribution has capacitance \[ C_{\mathrm{ext}}=\frac{Q_{\mathrm{outer}}}{V} =4\pi\varepsilon_0R. \]

Both capacitors share the shell as one terminal. Their other terminals—the ball and Earth—are connected by the grounding wire, so they are in parallel: \[ C=C_{\mathrm{gap}}+C_{\mathrm{ext}} =4\pi\varepsilon_0\left(\frac{rR}{R-r}+R\right) =\frac{4\pi\varepsilon_0R^2}{R-r}. \]

To draw the scheme, mark a node \(S\) for the outer shell and split the connection into two branches. Put one capacitor in each branch, labelled \(C_{\mathrm{gap}}=22.3\ \mathrm{pF}\) and \(C_{\mathrm{ext}}=22.3\ \mathrm{pF}\). Join their other ends at a common node with an Earth symbol. The potential difference across each capacitor is \(225\ \mathrm{V}\), with the shell terminal positive.

The solution was checked against the official solution.