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Two thin metal cups, each a hemisphere of radius 10 cm, are put together rim to rim to make a…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Two thin metal cups, each a hemisphere of radius $10\ \text{cm}$ , are put together rim to rim to make a closed spherical shell. The lower cup is fixed on an insulating stand. The upper cup, of mass $0.80\ \text{g}$ , simply rests on it, touching it all round the rim, as in the figure. The shell is given a charge of $+0.20\ \mu\text{C}$ . The charge spreads over the outside of the shell, and the cups stay in contact the whole time, so ignore the thin line where they meet. Treat the cups as one thin conducting sphere, and take $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$ and $g = 9.8\ \text{m/s}^2$ .

Two thin metal hemispherical cups of radius 10 cm, put rim to rim to make a sphere. The lower cup is fixed on an insulating stand. The upper cup, of mass 0.80 g, rests on the lower one. Small plus signs all round the outside show the charge of plus 0.20 microcoulombs on the shell.

What is the surface charge density on the shell?
What is the electric field just outside the shell?
A small patch of the shell feels the field of all the rest of the charge. What is that field?
What is the outward force on each square metre of the shell?
With what force does the charge push the upper cup away from the lower one?
The charge is now raised slowly. At what charge does the upper cup just lift off?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Find the field that the shell's own charge feels, then the pressure, then add up its upward parts over the upper cup.

Step 1. Density (a). $\sigma = \dfrac{Q}{4\pi R^2} = \dfrac{0.20 \times 10^{-6}\ \text{C}}{4\pi(0.10\ \text{m})^2} = 1.59\ \mu\text{C/m}^2$ .

Step 2. Field outside (b). $E = \dfrac{\sigma}{\varepsilon_0} = 1.80 \times 10^{5}\ \text{N/C}$ . Inside the metal and inside the shell it is zero.

Step 3. Field on a patch (c). A small patch alone gives $\sigma/(2\varepsilon_0)$ on each side, outwards on both. The rest of the shell must give the same $\sigma/(2\varepsilon_0)$ outwards at the patch, so that the total is $\sigma/\varepsilon_0$ outside and zero inside. The patch feels only the rest: $\dfrac{\sigma}{2\varepsilon_0} = 8.99 \times 10^{4}\ \text{N/C}$ .

Step 4. Pressure (d). $\dfrac{F}{A} = \sigma \times \dfrac{\sigma}{2\varepsilon_0} = \dfrac{\sigma^2}{2\varepsilon_0} = 0.143\ \text{N/m}^2$ , outwards.

Step 5. Push on the upper cup (e). Add up the upward parts of the pressure over the cup. The sideways parts cancel, and the upward parts add up to the pressure times the cup's shadow on the rim plane:

\displaystyle F = \pi R^2 \cdot \frac{\sigma^2}{2\varepsilon_0} = \frac{Q^2}{32\pi\varepsilon_0R^2} = 4.50\ \text{mN}

Step 6. Lift-off (f). The cup weighs $mg = 7.84\ \text{mN}$ , more than the push in (e), so it still rests. It lifts when $\dfrac{Q^2}{32\pi\varepsilon_0R^2} = mg$ :

\displaystyle Q = \sqrt{32\pi\varepsilon_0R^2mg} = 0.264\ \mu\text{C}

Answer: (a) $1.59\ \mu\text{C/m}^2$ . (b) $1.80 \times 10^{5}\ \text{N/C}$ . (c) $8.99 \times 10^{4}\ \text{N/C}$ . (d) $0.143\ \text{N/m}^2$ . (e) $4.50\ \text{mN}$ . (f) $0.264\ \mu\text{C}$ .

Watch out. Using the full field $\sigma/\varepsilon_0$ on the charge, and the curved area $2\pi R^2$ in place of $\pi R^2$ , gives $18.0\ \text{mN}$ , four times too much. Each slip is a factor of two.

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