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Three thin metal spherical shells A, B and C share the same centre, as in the figure…

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Three thin metal spherical shells $A$ , $B$ and $C$ share the same centre, as in the figure. Their radii are $5.0\ \text{cm}$ , $10.0\ \text{cm}$ and $15.0\ \text{cm}$ . Shell $A$ carries $+6.0\ \text{nC}$ , shell $B$ carries $-4.0\ \text{nC}$ and shell $C$ carries $+2.0\ \text{nC}$ . The shells are held apart by insulating supports, and nothing else is nearby. Each shell is so thin that its charge can be taken to sit on a sphere of its own radius. Potentials are measured from zero at infinity, which is also the potential of the earth. Give every potential and charge with its sign. Take $\dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^{9}\ \text{N m}^2\,\text{C}^{-2}$ .

Three thin concentric spherical shells seen in cross-section, with a common centre. The inner shell A has a radius of 5.0 cm and carries plus 6.0 nanocoulombs. The middle shell B has a radius of 10 cm and carries minus 4.0 nanocoulombs. The outer shell C has a radius of 15 cm and carries plus 2.0 nanocoulombs.

What is the potential of shell $A$ ?
What is the potential of shell $B$ ?

Now a thin insulated wire is passed through tiny holes in $C$ and $B$ . It joins shell $A$ to the earth, and it stays there. The charges on $B$ and $C$ do not change.

What is the charge on shell $A$ now?
How much charge flows from shell $A$ to the earth along the wire?
What is the potential of shell $B$ now?
What is the potential of shell $C$ now?

The worked solution

Have you tried it?

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. At any radius, each shell gives $kq/R$ if the point is inside it and $kq/r$ if the point is outside it. Add the three. Earthing $A$ fixes its potential at zero, and its charge adjusts.

Step 1. Shell $A$ (a). $A$ is inside all three shells:

\displaystyle V_A = k\left(\frac{6.0}{0.05} + \frac{-4.0}{0.10} + \frac{2.0}{0.15}\right) \times 10^{-9}\ \text{V} = 840\ \text{V}

Step 2. Shell $B$ (b). $B$ is outside $A$ and inside $C$ : $V_B = k\left(\dfrac{6.0 -4.0}{0.10} + \dfrac{2.0}{0.15}\right) \times 10^{-9}\ \text{V} = 300\ \text{V}$ .

Step 3. Earthing $A$ (c). Now $V_A = 0$ :

\displaystyle \frac{q_A}{0.05} + \frac{-4.0}{0.10} + \frac{2.0}{0.15} = 0 \quad\Rightarrow\quad q_A = 1.33\ \text{nC}

It is still positive. The two outer shells hold $A$ at a negative potential by themselves, so $A$ must keep some positive charge to reach zero.

Step 4. Charge to the earth (d). $6.00\ \text{nC} - 1.33\ \text{nC} = 4.67\ \text{nC}$ .

Step 5. The new potentials (e, f). $V_B = k\left(\dfrac{1.333 -4.0}{0.10} + \dfrac{2.0}{0.15}\right) \times 10^{-9}\ \text{V} = -120\ \text{V}$ , and $V_C = \dfrac{k(1.333 -4.0 +2.0) \times 10^{-9}}{0.15}\ \text{V} = -40\ \text{V}$ .

Answer: (a) $840\ \text{V}$ . (b) $300\ \text{V}$ . (c) $1.33\ \text{nC}$ . (d) $4.67\ \text{nC}$ . (e) $-120\ \text{V}$ . (f) $-40\ \text{V}$ .

Watch out. Earthing does not empty a shell. Taking $q_A = 0$ would give $V_C = -120\ \text{V}$ and leave $A$ at a negative potential, not at the potential of the earth.

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