The Workbook

\(\Phi = EA\cos\theta\), where \(\theta\) is the angle between \(\vec{E}\) and the normal to the area. \(A = 0.2 \times 0.2 = 0.04\) m\(^2\).

1. \(\theta = 0\): \(\Phi = 500 \times 0.04 = 20\) N m\(^2\)/C.

2. \(\theta = 60^\circ\): \(\Phi = 20 \times \dfrac{1}{2} = 10\) N m\(^2\)/C.

3. Field parallel to the square means \(\theta = 90^\circ\): \(\Phi = 0\). No field line crosses the square.

\(\Phi = \vec{E}\cdot\vec{A}\), with \(\vec{A} = 2\hat{n}\).

1. \(\hat{n} = \hat{i}\): \(\Phi = 3 \times 2 = 6\) N m\(^2\)/C.

2. \(\hat{n} = \hat{j}\): \(\Phi = 4 \times 2 = 8\) N m\(^2\)/C.

3. \(\hat{n} = \hat{k}\): \(\vec{E}\) has no \(z\)-component, so \(\Phi = 0\).

1. Gauss's law: \(\Phi = \dfrac{q_{\text{in}}}{\varepsilon_0} = \dfrac{(2 - 1) \times 10^{-6}}{8.85 \times 10^{-12}} \approx 1.13 \times 10^5\) N m\(^2\)/C.

2. The outside charge does not change the flux: every line from it that enters the surface also leaves. But it does change \(\vec{E}\) at points on the surface, since \(\vec{E}\) is the sum of the fields of all charges.

1. Total flux \(= \dfrac{q}{\varepsilon_0} = 6000\) N m\(^2\)/C. By symmetry the six faces share it equally: \(1000\) N m\(^2\)/C per face.

2. Eight identical cubes fit around a corner, and the charge sends equal flux into each. The flux through one cube is \(\dfrac{1}{8} \times 6000 = 750\) N m\(^2\)/C.

3. The three faces touching the charge get no flux (the field is along them). The other three share \(750\) equally: \(250\) N m\(^2\)/C each.

1. Flux is shared in proportion to solid angle: \(\Phi = \dfrac{q}{\varepsilon_0} \times \dfrac{\Omega}{4\pi}\).

2. \(\Omega = 2\pi(1 - \cos 60^\circ) = \pi\). \(\Phi = 6000 \times \dfrac{\pi}{4\pi} = 1500\) N m\(^2\)/C.

1. The slant distance from \(q\) to the rim is \(\sqrt{4^2 + 3^2} = 5\) cm. \(\cos\theta = \dfrac{d}{5} = \dfrac{4}{5} = 0.8\).

2. \(\Omega = 2\pi(1 - 0.8) = 0.4\pi\). \(\Phi = 6000 \times \dfrac{0.4\pi}{4\pi} = 6000 \times 0.1 = 600\) N m\(^2\)/C.

1. The enclosed charge is \(+q - q = 0\), so \(\Phi = 0\).

2. The field is not zero on the surface. Field lines leave \(+q\), and every line that crosses the surface outward comes back in to end on \(-q\). Outward and inward flux cancel.

1. The closed hemisphere encloses no charge, so the total flux is \(0\).

2. The field enters through the flat face, against its outward normal. \(\Phi_{\text{flat}} = -EA = -500 \times \pi(0.1)^2 = -5\pi \approx -15.7\) N m\(^2\)/C.

3. \(\Phi_{\text{flat}} + \Phi_{\text{curved}} = 0\), so \(\Phi_{\text{curved}} = +5\pi \approx 15.7\) N m\(^2\)/C. Every line that enters the flat face leaves through the curved face.

1. By symmetry \(\vec{E}\) points radially outward from the wire and has the same size at every point at distance \(r\).

   Take a cylinder of radius \(r\) and length \(L\) around the wire. No flux crosses the flat ends (\(\vec{E}\) is parallel to them). Through the curved surface, \(\Phi = E \times 2\pi rL\).

   Enclosed charge \(= \lambda L\). Gauss's law: \(E \cdot 2\pi rL = \dfrac{\lambda L}{\varepsilon_0}\), so \(E = \dfrac{\lambda}{2\pi\varepsilon_0 r} = \dfrac{2k\lambda}{r}\).

2. \(E = \dfrac{2 \times 9 \times 10^9 \times 5 \times 10^{-6}}{0.1} = 9 \times 10^5\) N/C.

1. By symmetry \(\vec{E}\) is perpendicular to the sheet, pointing away on both sides, with the same size on each side.

   Take a pillbox of face area \(A\) with one face on each side of the sheet. No flux crosses its curved side. Each flat face gives \(EA\), total \(2EA\).

   Enclosed charge \(= \sigma A\). So \(2EA = \dfrac{\sigma A}{\varepsilon_0}\) and \(E = \dfrac{\sigma}{2\varepsilon_0}\).

2. \(E = \dfrac{8.85 \times 10^{-9}}{2 \times 8.85 \times 10^{-12}} = 500\) N/C. The distance never entered the argument. The field lines are parallel and do not spread out, so the flux through the faces is the same at any distance.

1. A sphere of radius \(5\) cm encloses no charge, so \(E = 0\) inside the shell.

2. Just outside, the shell acts as a point charge at its centre: \(E = \dfrac{kQ}{R^2} = \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{(0.1)^2} = 1.8 \times 10^6\) N/C.

3. At \(20\) cm: \(E = \dfrac{kQ}{r^2} = \dfrac{1.8 \times 10^4}{0.04} = 4.5 \times 10^5\) N/C.

1. Take a Gaussian sphere of radius \(r < R\). The charge inside it is \(Q\dfrac{r^3}{R^3}\), since the charge is spread uniformly and volume scales as \(r^3\).

   Gauss's law: \(E \cdot 4\pi r^2 = \dfrac{Qr^3}{\varepsilon_0 R^3}\), so \(E = \dfrac{Qr}{4\pi\varepsilon_0 R^3} = \dfrac{kQr}{R^3}\).

2. At \(r = 5\) cm: \(E = \dfrac{9 \times 10^9 \times 2 \times 10^{-6} \times 0.05}{(0.1)^3} = \dfrac{900}{10^{-3}} = 9 \times 10^5\) N/C.

   At the surface: \(E = \dfrac{kQ}{R^2} = 1.8 \times 10^6\) N/C. At \(r = 20\) cm the sphere acts as a point charge: \(E = 4.5 \times 10^5\) N/C.

1. The graph rises in a straight line up to \(r = 10\) cm and then falls. The peak is at the surface, so \(R = 10\) cm.

2. At the surface \(E = \dfrac{kQ}{R^2}\), so \(Q = \dfrac{E R^2}{k} = \dfrac{1.8 \times 10^6 \times 0.01}{9 \times 10^9} = 2 \times 10^{-6}\) C \(= 2\,\mu\)C.

3. At \(5\) cm, half way up the straight line: \(E = 9 \times 10^5\) N/C. At \(20\) cm, twice the radius: \(E = \dfrac{1.8 \times 10^6}{4} = 4.5 \times 10^5\) N/C. Both agree with Question 12.

1. Charge inside radius \(r\): \(q_{\text{in}} = \rho \times \dfrac{4}{3}\pi r^3\).

   Gauss's law: \(E \cdot 4\pi r^2 = \dfrac{\rho \frac{4}{3}\pi r^3}{\varepsilon_0}\), so \(E = \dfrac{\rho r}{3\varepsilon_0}\).

2. \(\rho = \dfrac{2 \times 10^{-6}}{\frac{4}{3}\pi(0.1)^3} = \dfrac{2 \times 10^{-6}}{4.19 \times 10^{-3}} \approx 4.8 \times 10^{-4}\) C/m\(^3\).

   \(E = \dfrac{4.8 \times 10^{-4} \times 0.05}{3 \times 8.85 \times 10^{-12}} \approx 9 \times 10^5\) N/C, the same as Question 12.

Inside: \(E\left(\dfrac{R}{2}\right) = \dfrac{kQ(R/2)}{R^3} = \dfrac{kQ}{2R^2}\). Outside: \(E(2R) = \dfrac{kQ}{(2R)^2} = \dfrac{kQ}{4R^2}\).

Ratio \(= \dfrac{kQ/2R^2}{kQ/4R^2} = 2\).

\(E\) grows with \(r\) inside and falls with \(r\) outside, so it is largest at the surface, \(r = R\).

For each point draw a Gaussian sphere through it and count the charge inside.

\(P\) (\(5\) cm): no charge inside. \(E = 0\).

\(Q\) (\(15\) cm): only the inner shell, \(+2\,\mu\)C, is inside. \(E = \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{(0.15)^2} = \dfrac{1.8 \times 10^4}{0.0225} = 8 \times 10^5\) N/C, outward.

\(S\) (\(30\) cm): both shells inside, total charge \(0\). \(E = 0\).

\(E = 500\) N/C, pointing away from the sheet. The electron is negative, so the force on it points towards the sheet.

\(a = \dfrac{eE}{m_e} = \dfrac{1.6 \times 10^{-19} \times 500}{9.1 \times 10^{-31}} \approx 8.8 \times 10^{13}\) m/s\(^2\), towards the sheet.

1. Both spheres enclose the same charge, so the flux is the same: ratio \(1\).

2. \(E \propto \dfrac{1}{r^2}\), so the field on the smaller sphere is \(4\) times larger: ratio \(4\). The larger sphere has \(4\) times the area, which is why the flux comes out equal.

Curved surface: \(\vec{E}\) is parallel to it, so the flux is \(0\).

Flat faces: the field enters one face (\(\Phi = -EA\)) and leaves the other (\(\Phi = +EA\)).

Whole cylinder: \(0 - EA + EA = 0\), as Gauss's law demands for a surface with no charge inside.

1. No. The flux depends only on the charge enclosed, and the charge is still inside.

2. Yes. The field of a point charge depends on the distance from it, so moving the charge changes \(\vec{E}\) at each point of the surface. Only the total flux stays fixed.

Build a cube of side \(a\) with the square as one face. The charge is at height \(\dfrac{a}{2}\) above the centre of that face, which is the centre of the cube.

The flux through the whole cube is \(\dfrac{q}{\varepsilon_0} = 6000\) N m\(^2\)/C, shared equally by six faces. Through the square: \(1000\) N m\(^2\)/C.

1. Charge is proportional to volume, and volume is proportional to \(r^3\). Inside \(\dfrac{R}{2}\): \(\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}\) of \(Q\).

2. Gauss's law on a sphere of radius \(\dfrac{R}{2}\): \(E \cdot 4\pi\left(\dfrac{R}{2}\right)^2 = \dfrac{Q/8}{\varepsilon_0}\), so \(E = \dfrac{Q/8}{4\pi\varepsilon_0 R^2/4} = \dfrac{kQ}{2R^2}\).