The Workbook

1. A metal has free electrons. If there were a field inside, the electrons would feel a force and move. In electrostatics nothing moves, so the field inside must be zero.

2. Take a Gaussian surface just inside the metal surface. \(E = 0\) on it, so the flux is zero and the enclosed charge is zero. Any charge on the body must therefore lie on the surface.

1. \(E = 0\) inside, so no work is done in moving a charge between any two points of the metal. No work means no potential difference: the whole body is one equipotential.

2. A field component along the surface would push the free surface charges along the surface. They would move, which does not happen in electrostatics. So the field just outside has no component along the surface.

1. Take a small pillbox of face area \(A\) with one face just outside the surface and the other inside the metal.

   Inside, \(E = 0\), so that face gives no flux. The curved side gives none, since \(\vec{E}\) is perpendicular to the surface. Only the outer face gives \(EA\).

   Enclosed charge \(= \sigma A\). Gauss: \(EA = \dfrac{\sigma A}{\varepsilon_0}\), so \(E = \dfrac{\sigma}{\varepsilon_0}\).

2. \(E = \dfrac{8.85 \times 10^{-9}}{8.85 \times 10^{-12}} = 1000\) N/C. A thin sheet sends its flux out on both sides, so each side gets \(\dfrac{\sigma}{2\varepsilon_0}\). A conductor sends all the flux out on one side, because the field inside is zero.

1. Inside the metal (centre and \(5\) cm): \(E = 0\).

   Just outside: \(E = \dfrac{kQ}{R^2} = \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{(0.1)^2} = 1.8 \times 10^6\) N/C. At \(20\) cm: \(E = \dfrac{1.8 \times 10^4}{0.04} = 4.5 \times 10^5\) N/C.

2. Surface: \(V = \dfrac{kQ}{R} = \dfrac{1.8 \times 10^4}{0.1} = 1.8 \times 10^5\) V. The centre is at the same potential, since \(E = 0\) inside.

1. A Gaussian surface inside the metal has \(E = 0\), so it encloses zero charge. The inner surface must carry \(-q = -2\,\mu\)C to cancel \(q\).

   The shell is neutral, so the outer surface carries \(+2\,\mu\)C.

2. \(P\) (cavity, \(2\) cm): only \(q\) is inside a sphere through \(P\). \(E = \dfrac{kq}{r^2} = \dfrac{1.8 \times 10^4}{(0.02)^2} = 4.5 \times 10^7\) N/C.

   \(Q\) (metal): \(E = 0\).

   \(S\) (\(20\) cm): total enclosed charge \(q - q + q = 2\,\mu\)C. \(E = \dfrac{1.8 \times 10^4}{0.04} = 4.5 \times 10^5\) N/C.

1. No. The outer surface still carries \(+2\,\mu\)C, spread uniformly, because the metal screens the outside from whatever happens in the cavity. The outside field stays \(\dfrac{kq}{r^2}\).

2. No. The inner surface charge crowds towards the point nearest \(q\), so that its field cancels the field of \(q\) everywhere inside the metal.

3. Earthing lets the \(+2\,\mu\)C on the outer surface flow to the earth. The outer surface charge becomes \(0\), and the field outside becomes \(0\). The inner surface still carries \(-2\,\mu\)C.

Inside a closed conductor, the field due to outside charges is zero. The charges induced on the conductor's surface arrange themselves to cancel the outside field exactly.

The metal box blocks the electric field of the radio signal, so the phone receives nothing. The car body carries the lightning current round the outside; the field inside stays zero and the person is safe.

1. The field pushes free electrons opposite to \(\vec{E}\), that is, to the left. Negative charge collects on the left face and positive charge on the right face.

2. The induced charges produce a field that cancels \(\vec{E}\) inside the metal: \(E = 0\) there. The cavity is surrounded by metal with no charge inside it, so the field in the cavity is also \(0\).

A Gaussian surface inside the metal of \(B\) encloses zero charge. It contains \(A\) (\(+2\,\mu\)C) and the inner surface of \(B\). So the inner surface of \(B\) carries \(-2\,\mu\)C.

\(B\) carries \(+4\,\mu\)C in total, so its outer surface carries \(4 - (-2) = +6\,\mu\)C.

The potential at a point is the sum of the potentials of all the shells. A shell contributes \(\dfrac{kQ}{r}\) outside and \(\dfrac{kQ}{R}\) inside.

\(V_A = \dfrac{kQ_A}{a} + \dfrac{kQ_B}{b} = 9 \times 10^9\left(\dfrac{2 \times 10^{-6}}{0.1} + \dfrac{4 \times 10^{-6}}{0.2}\right) = 1.8 \times 10^5 + 1.8 \times 10^5 = 3.6 \times 10^5\) V.

\(V_B = \dfrac{k(Q_A + Q_B)}{b} = \dfrac{9 \times 10^9 \times 6 \times 10^{-6}}{0.2} = 2.7 \times 10^5\) V.

1. Earthing makes \(V_B = 0\): \(\dfrac{k(Q_A + Q_B)}{b} = 0\), so \(Q_B = -Q_A = -2\,\mu\)C. The \(+6\,\mu\)C on the outer surface flows away; the \(-2\,\mu\)C on the inner surface stays.

2. \(V_A = \dfrac{kQ_A}{a} + \dfrac{kQ_B}{b} = 9 \times 10^9 \times 2 \times 10^{-6}\left(\dfrac{1}{0.1} - \dfrac{1}{0.2}\right) = 1.8 \times 10^4 \times 5 = 9 \times 10^4\) V.

Earthing \(A\) makes \(V_A = 0\), but \(A\) does not become neutral: charge flows until its potential is zero.

\(\dfrac{kQ_A}{a} + \dfrac{kQ_B}{b} = 0\), so \(Q_A = -Q_B\dfrac{a}{b} = -4 \times \dfrac{0.1}{0.2} = -2\,\mu\)C.

1. Joined shells form one conductor, and all the charge of a conductor sits on its outer surface. \(A\) ends with \(0\) and \(B\) with the total, \(6\,\mu\)C.

2. \(V = \dfrac{k \times 6 \times 10^{-6}}{0.2} = 2.7 \times 10^5\) V, the same for both shells.

\(V_A = \dfrac{kQ}{a} + \dfrac{kQ_B}{b}\) and \(V_B = \dfrac{kQ}{b} + \dfrac{kQ_B}{b}\).

Subtracting, the \(Q_B\) terms cancel: \(V_A - V_B = kQ\left(\dfrac{1}{a} - \dfrac{1}{b}\right)\). The outer shell shifts both potentials by the same amount.

\(V_A - V_B = 9 \times 10^9 \times 2 \times 10^{-6}\left(10 - 5\right) = 9 \times 10^4\) V.

1. \(Q = \dfrac{VR}{k}\). Small sphere: \(Q_1 = \dfrac{9000 \times 0.1}{9 \times 10^9} = 10^{-7}\) C \(= 100\) nC. Large sphere: \(Q_2 = 200\) nC.

2. \(\sigma = \dfrac{Q}{4\pi R^2}\), and with \(V\) fixed, \(Q \propto R\), so \(\sigma \propto \dfrac{1}{R}\). \(\dfrac{\sigma_1}{\sigma_2} = \dfrac{R_2}{R_1} = 2\). The smaller sphere has the denser charge.

1. Charge flows until both spheres are at the same potential: \(\dfrac{kQ_1}{R_1} = \dfrac{kQ_2}{R_2}\), so \(\dfrac{Q_1}{Q_2} = \dfrac{R_1}{R_2} = \dfrac{1}{2}\).

   \(Q_1 + Q_2 = 6\,\mu\)C, so \(Q_1 = 2\,\mu\)C and \(Q_2 = 4\,\mu\)C.

2. \(V = \dfrac{kQ_1}{R_1} = \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{0.1} = 1.8 \times 10^5\) V.

1. \(E = \dfrac{\sigma}{\varepsilon_0}\), so \(\sigma = \varepsilon_0 E = 8.85 \times 10^{-12} \times 10^6 = 8.85 \times 10^{-6}\) C/m\(^2\).

2. \(P = \dfrac{\sigma^2}{2\varepsilon_0} = \dfrac{(8.85 \times 10^{-6})^2}{2 \times 8.85 \times 10^{-12}} \approx 4.4\) N/m\(^2\). Equivalently \(P = \dfrac{1}{2}\varepsilon_0 E^2\).

Just outside a small patch, the total field is \(\dfrac{\sigma}{\varepsilon_0}\). Just inside it is \(0\). The patch itself, like a small sheet, gives \(\dfrac{\sigma}{2\varepsilon_0}\) outward on the outside and \(\dfrac{\sigma}{2\varepsilon_0}\) inward on the inside.

So the rest of the conductor must give \(\dfrac{\sigma}{2\varepsilon_0}\) outward on both sides of the patch. This is the field that acts on the patch's own charge.

Force on a patch of area \(dA\): \(dF = (\sigma\,dA)\dfrac{\sigma}{2\varepsilon_0}\). Pressure \(= \dfrac{dF}{dA} = \dfrac{\sigma^2}{2\varepsilon_0}\), outward.

The outward pressure on each patch is the same, but the patches point in different directions. The net force on a hemisphere equals \(P\) times its projected area, a disc of radius \(R\).

\(F = P\pi R^2 = 4.4 \times \pi \times (0.1)^2 \approx 0.14\) N, pushing the two hemispheres apart.

1. Balance: \(\dfrac{\sigma^2}{2\varepsilon_0} = \dfrac{4T}{r} = \dfrac{4 \times 0.03}{0.01} = 12\) N/m\(^2\).

   \(\sigma = \sqrt{2\varepsilon_0 \times 12} = \sqrt{2 \times 8.85 \times 10^{-12} \times 12} \approx 1.46 \times 10^{-5}\) C/m\(^2\).

2. \(Q = \sigma \times 4\pi r^2 = 1.46 \times 10^{-5} \times 4\pi \times 10^{-4} \approx 1.8 \times 10^{-8}\) C \(= 18\) nC.

1. The field is largest just outside the surface: \(\dfrac{kQ}{R^2} = 3 \times 10^6\). \(Q_{\max} = \dfrac{3 \times 10^6 \times 0.01}{9 \times 10^9} \approx 3.3 \times 10^{-6}\) C \(= 3.3\,\mu\)C.

2. \(V_{\max} = \dfrac{kQ_{\max}}{R} = E_{\max}R = 3 \times 10^6 \times 0.1 = 3 \times 10^5\) V. A bigger sphere can hold a higher potential.

1. Outside the shell the total enclosed charge is \(q\). So the field outside is \(\dfrac{kq}{r^2}\) and the potential of the outer surface is \(\dfrac{kq}{b}\).

   \(V_{\text{shell}} = \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{0.1} = 1.8 \times 10^5\) V. The whole shell is at this potential.

2. Three charges contribute: \(q\) itself, \(-q\) on the inner surface at radius \(a\), and \(+q\) on the outer surface at radius \(b\). So \(V = \dfrac{kq}{r} - \dfrac{kq}{a} + \dfrac{kq}{b}\).

   \(V = 1.8 \times 10^4\left(\dfrac{1}{0.025} - \dfrac{1}{0.05} + \dfrac{1}{0.1}\right) = 1.8 \times 10^4 \times 30 = 5.4 \times 10^5\) V.