The Workbook

1. \(V = \dfrac{W}{q} = \dfrac{4 \times 10^{-3}}{2 \times 10^{-6}} = 2000\) V.

2. \(W = qV = (-1 \times 10^{-6})(2000) = -2 \times 10^{-3}\) J. The field pulls the negative charge in, so the agent does negative work.

1. \(V = \dfrac{kq}{r}\). At \(20\) cm: \(V = \dfrac{9 \times 10^9 \times 4 \times 10^{-6}}{0.2} = 1.8 \times 10^5\) V.

   At \(40\) cm the distance is doubled, so \(V\) is halved: \(9 \times 10^4\) V.

2. Potential is a scalar; it has a value but no direction. At infinity \(V = 0\) by choice.

1. Each charge is \(10\) cm from the midpoint. \(V = \dfrac{k(2 \times 10^{-6})}{0.1} + \dfrac{k(-2 \times 10^{-6})}{0.1} = 0\). Potentials add as numbers, and these two cancel.

2. No. The fields of the two charges at the midpoint both point from \(+2\,\mu\)C towards \(-2\,\mu\)C, so they add.

   \(E = 2 \times \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{(0.1)^2} = 3.6 \times 10^6\) N/C. Zero potential does not mean zero field.

Between the charges, at \(x\) from \(+3\,\mu\)C: \(\dfrac{k(3)}{x} = \dfrac{k(2)}{0.3 - x}\), so \(0.9 - 3x = 2x\) and \(x = 0.18\) m.

Outside, the point must be nearer the smaller charge, at \(x\) from \(+3\,\mu\)C: \(\dfrac{3}{x} = \dfrac{2}{x - 0.3}\), so \(3x - 0.9 = 2x\) and \(x = 0.9\) m.

The two zero-potential points are \(18\) cm and \(90\) cm from the \(+3\,\mu\)C charge.

1. Each corner is half a diagonal, \(10\) cm, from the centre. \(V = 4 \times \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{0.1} = 7.2 \times 10^5\) V.

2. The fields of opposite corners are equal and opposite at the centre. \(E = 0\). Potentials add as numbers; fields add as vectors.

\(r = \sqrt{3^2 + 4^2} = 5\) m. \(V = \dfrac{kq}{r} = \dfrac{9 \times 10^9 \times 5 \times 10^{-6}}{5} = 9000\) V.

No. The potential of a point charge depends only on the distance \(r\), not on the direction.

1. \(W_{\text{ext}} = q(V_A - V_B) = 2 \times 10^{-6} \times (100 - 40) = 1.2 \times 10^{-4}\) J. Positive, since a positive charge is pushed to a higher potential.

2. The field does the opposite work: \(W_{\text{field}} = -1.2 \times 10^{-4}\) J.

1. \(KE = eV = 1.6 \times 10^{-19} \times 100 = 1.6 \times 10^{-17}\) J. In electron volts this is simply \(100\) eV.

2. \(\dfrac{1}{2}mv^2 = 1.6 \times 10^{-17}\), so \(v = \sqrt{\dfrac{2 \times 1.6 \times 10^{-17}}{9.1 \times 10^{-31}}} \approx 5.9 \times 10^6\) m/s.

1. \(\Delta U = q(V_A - V_B) = 2 \times 10^{-6} \times (50 - 200) = -3 \times 10^{-4}\) J. A positive charge loses potential energy when it moves to a lower potential.

2. Energy is conserved: \(KE_A = -\Delta U = 3 \times 10^{-4}\) J.

1. Outside and on the surface, the shell acts like a point charge at its centre. \(V = \dfrac{kQ}{R} = \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{0.1} = 1.8 \times 10^5\) V.

2. Inside the shell \(E = 0\), so the potential does not change from the surface to the centre. \(V = 1.8 \times 10^5\) V.

3. At \(20\) cm: \(V = \dfrac{kQ}{r} = \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{0.2} = 9 \times 10^4\) V.

1. Every element of the ring is a distance \(R\) from \(C\). \(V = \dfrac{kQ}{R} = \dfrac{9 \times 10^9 \times 5 \times 10^{-6}}{0.3} = 1.5 \times 10^5\) V.

2. Every element is a distance \(\sqrt{R^2 + x^2} = 0.5\) m from \(P\). \(V = \dfrac{kQ}{0.5} = 9 \times 10^4\) V.

3. Potential is a scalar, so the contributions of all elements add. Field is a vector, and at \(C\) the fields of opposite elements cancel in pairs.

1. Every element of the semicircle is \(R\) from the centre. \(V = \dfrac{kQ}{R} = \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{0.1} = 1.8 \times 10^5\) V.

2. The same: every element is still \(R\) from the centre. \(V = 1.8 \times 10^5\) V.

3. No. The potential at the centre depends only on the total charge and the radius, because every bit of charge is at the same distance.

1. Put \(P\) at the origin and the rod from \(x = a\) to \(x = a + L\). An element \(dx\) carries \(dq = \lambda\,dx\) and gives \(dV = \dfrac{k\lambda\,dx}{x}\).

   \(V = k\lambda\displaystyle\int_a^{a+L}\dfrac{dx}{x} = k\lambda\left[\ln x\right]_a^{a+L} = k\lambda\ln\dfrac{a + L}{a}\).

2. \(k\lambda = 9 \times 10^9 \times 2 \times 10^{-6} = 1.8 \times 10^4\) V. \(V = 1.8 \times 10^4 \times \ln\dfrac{0.4}{0.1} = 1.8 \times 10^4 \times 1.39 \approx 2.5 \times 10^4\) V.

\(\dfrac{\sigma}{2\varepsilon_0} = \dfrac{1.77 \times 10^{-8}}{2 \times 8.85 \times 10^{-12}} = 1000\) V/m.

1. At the centre \(x = 0\): \(V = 1000 \times (R - 0) = 1000 \times 0.3 = 300\) V.

2. \(\sqrt{x^2 + R^2} = \sqrt{0.16 + 0.09} = 0.5\) m. \(V = 1000 \times (0.5 - 0.4) = 100\) V.

1. \(E_x = -\dfrac{dV}{dx} = -5\) V/m. The field has magnitude \(5\) V/m and points along \(-x\), from higher potential to lower potential.

2. Yes. The derivative of \(5x + 3\) is the constant \(5\), so \(E\) is the same everywhere. The field is uniform.

\(E_x = -\dfrac{\partial V}{\partial x} = -4x\) and \(E_y = -\dfrac{\partial V}{\partial y} = 3\), in V/m.

At \((1, 2)\): \(E_x = -4\) V/m, \(E_y = 3\) V/m. \(|E| = \sqrt{16 + 9} = 5\) V/m.

1. \(E = -\dfrac{dV}{dr} = -\dfrac{d}{dr}\left(\dfrac{20}{r}\right) = \dfrac{20}{r^2}\) V/m. It is positive, so the field points radially outward: the charge is positive.

2. At \(r = 2\) m: \(E = \dfrac{20}{4} = 5\) V/m.

3. Compare \(V = \dfrac{kq}{r}\) with \(\dfrac{20}{r}\): \(kq = 20\), so \(q = \dfrac{20}{9 \times 10^9} \approx 2.2 \times 10^{-9}\) C \(= 2.2\) nC.

\(E_x = -\dfrac{dV}{dx}\) is the negative of the slope of the graph.

Region \(A\): \(V\) is constant, slope \(0\), so \(E_x = 0\).

Region \(B\): \(V\) falls from \(10\) V to \(-10\) V over \(2\) m. Slope \(= \dfrac{-20}{2} = -10\) V/m, so \(E_x = +10\) V/m. The field points along \(+x\), the direction in which \(V\) falls.

Region \(C\): constant \(V\), so \(E_x = 0\).

1. \(V(2) - V(0) = -\displaystyle\int_0^2 E_x\,dx = -(5 \times 2) = -10\) V. So \(V(2) = 20 - 10 = 10\) V. The integral is the area under the graph.

2. From \(2\) m to \(4\) m the field is zero, so the potential does not change. \(V(4) = 10\) V.

3. The potential falls along the field. Moving along \(E\) means \(dV = -E\,dx\) is negative.

1. In a uniform field, \(V_A - V_B = E\,d = 200 \times 0.5 = 100\) V. Potential falls along the field, so \(A\) is at the higher potential.

2. \(BC\) is perpendicular to \(E\), so no potential changes along it: \(V_B - V_C = 0\).

   \(V_A - V_C = (V_A - V_B) + (V_B - V_C) = 100 + 0 = 100\) V. Only the displacement along the field matters.

1. \(E = \dfrac{V}{d} = \dfrac{400}{0.02} = 2 \times 10^4\) V/m. The field points from the higher potential to the lower, so upward.

2. \(F = eE = 1.6 \times 10^{-19} \times 2 \times 10^4 = 3.2 \times 10^{-15}\) N. The electron is negative, so the force is opposite to \(E\): downward, towards the \(+400\) V plate.

3. The electron moves \(1\) cm along the force. \(W = F \times d = 3.2 \times 10^{-15} \times 0.01 = 3.2 \times 10^{-17}\) J. Check: \(W = e \times \Delta V = 1.6 \times 10^{-19} \times 200 = 3.2 \times 10^{-17}\) J.

For a point charge \(V = \dfrac{kq}{r}\) and \(E = \dfrac{kq}{r^2}\), so \(\dfrac{V}{E} = r\).

\(r = \dfrac{900}{9000} = 0.1\) m \(= 10\) cm.

\(q = \dfrac{Vr}{k} = \dfrac{900 \times 0.1}{9 \times 10^9} = 10^{-8}\) C \(= 10\) nC.

1. \(U = \dfrac{kq_1 q_2}{r} = \dfrac{9 \times 10^9 \times 2 \times 10^{-6} \times 3 \times 10^{-6}}{0.3} = \dfrac{0.054}{0.3} = 0.18\) J.

   Positive energy means work had to be done to push the two like charges together from infinity. Released, they would fly apart and gain this much kinetic energy.

2. \(U = \dfrac{9 \times 10^9 \times 2 \times 10^{-6} \times (-3 \times 10^{-6})}{0.3} = -0.18\) J. Unlike charges have negative energy; work is needed to pull them apart.

\(U_1 = 0.18\) J at \(30\) cm. At \(60\) cm the distance doubles, so \(U_2 = 0.09\) J.

\(W_{\text{ext}} = U_2 - U_1 = 0.09 - 0.18 = -0.09\) J. The repulsion pushes the charges apart on its own, so the agent does negative work.

\(U = \dfrac{9 \times 10^9 \times 4 \times 10^{-6} \times (-4 \times 10^{-6})}{0.1} = -1.44\) J.

At infinity \(U = 0\). \(W_{\text{ext}} = U_{\text{final}} - U_{\text{initial}} = 0 - (-1.44) = 1.44\) J.

1. \(U = qV = 2 \times 10^{-6} \times 500 = 10^{-3}\) J.

2. It is the energy of the system: the charge together with whatever charges produce the potential. A single charge alone has no interaction energy.

Initial energy: \(U = \dfrac{9 \times 10^9 \times 2 \times 10^{-6} \times 3 \times 10^{-6}}{0.3} = 0.18\) J, kinetic energy zero.

Far away \(U = 0\), so all of it becomes kinetic energy: \(\dfrac{1}{2}mv^2 = 0.18\).

\(v^2 = \dfrac{2 \times 0.18}{0.01} = 36\), so \(v = 6\) m/s.

1. Three charges make \(3\) pairs. Each pair: \(U = \dfrac{kq^2}{a} = \dfrac{9 \times 10^9 \times (2 \times 10^{-6})^2}{0.2} = 0.18\) J.

2. \(U = 3 \times 0.18 = 0.54\) J. The energy of a system is the sum over all pairs, each counted once.

Two pairs now contain a \(+2\,\mu\)C and a \(-2\,\mu\)C: each has \(U = -0.18\) J. One pair is still \(+0.18\) J.

\(U = 0.18 - 0.18 - 0.18 = -0.18\) J.

1. Side pair: \(\dfrac{kq^2}{a} = \dfrac{9 \times 10^9 \times 10^{-12}}{0.1} = 0.09\) J.

   Diagonal pair (distance \(a\sqrt{2}\)): \(\dfrac{kq^2}{a\sqrt{2}} = \dfrac{0.09}{1.414} \approx 0.064\) J.

2. There are \(4\) side pairs and \(2\) diagonal pairs. \(U = 4\dfrac{kq^2}{a} + 2\dfrac{kq^2}{a\sqrt{2}} = \dfrac{kq^2}{a}\left(4 + \dfrac{2}{\sqrt{2}}\right) = \dfrac{kq^2}{a}\left(4 + \sqrt{2}\right)\).

   \(U = 0.09 \times (4 + 1.414) \approx 0.49\) J.

1. \(U_{AB} = \dfrac{k(2)(-1) \times 10^{-12}}{0.1} = -0.18\) J. \(U_{BC} = \dfrac{k(-1)(2) \times 10^{-12}}{0.1} = -0.18\) J. \(U_{AC} = \dfrac{k(2)(2) \times 10^{-12}}{0.2} = 0.18\) J.

   \(U = -0.18 - 0.18 + 0.18 = -0.18\) J.

2. Removing \(C\) takes away the two pairs that contain \(C\): \(U_{AC} + U_{BC} = 0.18 - 0.18 = 0\).

   \(W_{\text{ext}} = U_{\text{final}} - U_{\text{initial}} = U_{AB} - U = -0.18 - (-0.18) = 0\). The attraction of \(B\) and the repulsion of \(A\) balance exactly.

1. \(U = \dfrac{kQ^2}{2R} = \dfrac{9 \times 10^9 \times (2 \times 10^{-6})^2}{2 \times 0.1} = \dfrac{0.036}{0.2} = 0.18\) J.

2. \(U \propto Q^2\), so doubling \(Q\) makes \(U\) four times: \(0.72\) J. The extra energy needed is \(0.72 - 0.18 = 0.54\) J.

1. \(U = \dfrac{3kQ^2}{5R} = \dfrac{3 \times 9 \times 10^9 \times 4 \times 10^{-12}}{5 \times 0.1} = \dfrac{0.108}{0.5} = 0.216\) J.

2. In the solid sphere the charge is spread through the volume, so on average the bits of charge are closer together than on the shell. More work is needed to assemble them.

1. Every bit of the shell's charge is a distance \(R\) from the centre. \(U_{\text{int}} = \dfrac{kQq}{R} = \dfrac{9 \times 10^9 \times 2 \times 10^{-6} \times 10^{-6}}{0.1} = 0.18\) J.

2. Total \(= \) self-energy of the shell \(+\) interaction energy \(= 0.18 + 0.18 = 0.36\) J. A point charge has no finite self-energy, so it is not counted.

1. Spheres centred on the charge, since \(V = \dfrac{kq}{r}\) depends only on \(r\).

2. Planes perpendicular to the field, since \(V\) changes only along the field.

3. Planes parallel to the sheet. The field of a large sheet is uniform and perpendicular to the sheet.

1. Moving a charge along the surface changes no potential, so the field does no work. Work is zero only if the force is perpendicular to the displacement. So \(E\) is perpendicular to the surface.

2. A point has only one value of potential. If two surfaces with different potentials crossed, the crossing point would have two potentials at once.

1. \(W_{\text{field}} = q(V_{\text{initial}} - V_{\text{final}}) = q \times 0 = 0\). The distance does not matter.

2. Zero. The electrostatic force is conservative: the work depends only on the end points, and here they are the same point.

1. \(E = \dfrac{\Delta V}{d} = \dfrac{10}{0.02} = 500\) V/m. The field points from high potential to low, from the \(30\) V surface towards \(Q\).

2. \(W_{\text{field}} = q(V_P - V_Q) = (-2 \times 10^{-6})(30 - 0) = -6 \times 10^{-5}\) J. The negative charge is pushed against its motion, so the field does negative work.

1. \(V = \dfrac{kq}{r}\), so \(q = \dfrac{Vr}{k} = \dfrac{900 \times 0.1}{9 \times 10^9} = 10^{-8}\) C \(= 10\) nC. Check: at \(20\) cm \(V\) halves to \(450\) V.

2. \(W_{\text{ext}} = q(V_{\text{final}} - V_{\text{initial}}) = 2 \times 10^{-6} \times (900 - 450) = 9 \times 10^{-4}\) J.

3. No. The work depends only on the potentials of the two surfaces. It does not depend on the path between them.

1. Every point of the bisecting plane is equidistant from \(+q\) and \(-q\). The two potentials cancel: \(V = 0\) everywhere on the plane.

2. The plane is an equipotential surface, so \(W = q\,\Delta V = 0\).

Far away \(U = 0\) and \(KE = \dfrac{1}{2} \times 0.01 \times 36 = 0.18\) J.

At the closest point the particle stops for an instant, so all the kinetic energy has become potential energy: \(\dfrac{kq_1 q_2}{r} = 0.18\).

\(r = \dfrac{9 \times 10^9 \times 2 \times 10^{-6} \times 2 \times 10^{-6}}{0.18} = \dfrac{0.036}{0.18} = 0.2\) m \(= 20\) cm.

At closest approach the alpha particle stops and \(KE = \dfrac{k(2e)(79e)}{r}\).

\(KE = 5 \times 1.6 \times 10^{-13} = 8 \times 10^{-13}\) J.

\(r = \dfrac{9 \times 10^9 \times 158 \times (1.6 \times 10^{-19})^2}{8 \times 10^{-13}} = \dfrac{3.64 \times 10^{-26}}{8 \times 10^{-13}} \approx 4.5 \times 10^{-14}\) m.

1. In a uniform field \(V_A - V_B = Ed = 2000 \times 1 = 2000\) V. \(A\) is at the higher potential.

2. \(W_{\text{field}} = q(V_A - V_B) = 2 \times 10^{-6} \times 2000 = 4 \times 10^{-3}\) J. Check: \(W = qEd = 2 \times 10^{-6} \times 2000 \times 1\).

   \(\dfrac{1}{2}mv^2 = 4 \times 10^{-3}\), so \(v^2 = \dfrac{2 \times 4 \times 10^{-3}}{0.008} = 1\) and \(v = 1\) m/s.

\(U = \dfrac{k(e)(-e)}{r} = -\dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{5.3 \times 10^{-11}} = -\dfrac{2.304 \times 10^{-28}}{5.3 \times 10^{-11}} \approx -4.35 \times 10^{-18}\) J.

In electron volts: \(\dfrac{-4.35 \times 10^{-18}}{1.6 \times 10^{-19}} \approx -27\) eV. The negative sign shows the electron is bound to the proton.

1. \(p = q \times 2a = 2 \times 10^{-6} \times 2 \times 10^{-3} = 4 \times 10^{-9}\) C m.

2. \(\vec{p}\) points from the negative charge to the positive charge.

\(p = e\,d\), so \(d = \dfrac{p}{e} = \dfrac{3.4 \times 10^{-30}}{1.6 \times 10^{-19}} \approx 2.1 \times 10^{-11}\) m. This is about one fifth of the size of an atom.

Dipole moments add as vectors. For two equal vectors at angle \(\phi\), the resultant is \(2p\cos\dfrac{\phi}{2}\).

1. \(\phi = 0\): \(2p\).

2. \(\phi = 60^\circ\): \(2p\cos 30^\circ = \sqrt{3}\,p\).

3. \(\phi = 180^\circ\): \(0\). The two dipoles cancel.

Axial field: \(E = \dfrac{2kp}{r^3} = \dfrac{2 \times 9 \times 10^9 \times 4 \times 10^{-9}}{(0.1)^3} = \dfrac{72}{10^{-3}} = 7.2 \times 10^4\) N/C.

On the axis the field points along \(\vec{p}\), on both sides of the dipole.

Equatorial field: \(E = \dfrac{kp}{r^3} = \dfrac{9 \times 10^9 \times 4 \times 10^{-9}}{10^{-3}} = 3.6 \times 10^4\) N/C, directed opposite to \(\vec{p}\).

Ratio: \(\dfrac{E_{\text{axial}}}{E_{\text{eq}}} = \dfrac{2kp/r^3}{kp/r^3} = 2\).

For a short dipole \(E \propto \dfrac{1}{r^3}\).

1. At \(2r\): \(E \to \dfrac{E}{8}\).

2. At \(\dfrac{r}{2}\): \(E \to 8E\).

A point charge's field falls as \(\dfrac{1}{r^2}\), so doubling \(r\) only quarters it. The dipole field falls faster because its two charges nearly cancel from far away.

1. \(\theta = 0\): \(E = \dfrac{kp}{r^3}\sqrt{1 + 3} = \dfrac{2kp}{r^3}\), the axial field.

   \(\theta = 90^\circ\): \(E = \dfrac{kp}{r^3}\sqrt{1 + 0} = \dfrac{kp}{r^3}\), the equatorial field.

2. \(\tan\alpha = \dfrac{1}{2}\tan 45^\circ = \dfrac{1}{2}\). The field is tilted less than \(OP\) is, towards the axis.

3. \(OP\) makes angle \(\theta\) with \(\vec{p}\), and \(\vec{E}\) makes a further angle \(\alpha\) with \(OP\). So \(\vec{E}\) makes angle \(\theta + \alpha\) with \(\vec{p}\).

   Perpendicular means \(\theta + \alpha = 90^\circ\), so \(\tan\alpha = \cot\theta\). With \(\tan\alpha = \dfrac{1}{2}\tan\theta\): \(\dfrac{1}{2}\tan\theta = \dfrac{1}{\tan\theta}\), so \(\tan^2\theta = 2\).

   \(\tan\theta = \sqrt{2}\), \(\theta \approx 55^\circ\). Not on the equatorial line: there \(\vec{E}\) is antiparallel to \(\vec{p}\), not perpendicular.

\(\dfrac{kp}{r^3} = \dfrac{9 \times 10^9 \times 4 \times 10^{-9}}{(0.1)^3} = 3.6 \times 10^4\) N/C.

\(A\) is axial: \(E = 2 \times 3.6 \times 10^4 = 7.2 \times 10^4\) N/C along \(\vec{p}\).

\(B\) is equatorial: \(E = 3.6 \times 10^4\) N/C, opposite to \(\vec{p}\).

\(C\) is axial on the other side: \(E = 7.2 \times 10^4\) N/C, again along \(\vec{p}\). On the axis the field points the same way on both sides.

Put the centre at the origin, \(+q\) at \(x = a\) and \(-q\) at \(x = -a\). At \(x = r\) on the axis, \(+q\) is a distance \((r - a)\) away and \(-q\) is \((r + a)\) away.

\(E = \dfrac{kq}{(r - a)^2} - \dfrac{kq}{(r + a)^2} = kq\,\dfrac{(r + a)^2 - (r - a)^2}{(r^2 - a^2)^2} = \dfrac{kq \cdot 4ar}{(r^2 - a^2)^2}\).

For \(r \gg a\), \(r^2 - a^2 \approx r^2\), so \(E \approx \dfrac{4kqar}{r^4} = \dfrac{2k(2aq)}{r^3} = \dfrac{2kp}{r^3}\), directed along \(\vec{p}\).

1. Axial: \(V = \dfrac{kp}{r^2} = \dfrac{9 \times 10^9 \times 4 \times 10^{-9}}{(0.1)^2} = \dfrac{36}{0.01} = 3600\) V.

2. Equatorial: \(V = 0\). Every point there is equidistant from \(+q\) and \(-q\).

\(V = \dfrac{kp\cos\theta}{r^2} = 3600 \times \cos 60^\circ = 3600 \times \dfrac{1}{2} = 1800\) V.

Potential: every point of the equatorial line is the same distance from \(+q\) and from \(-q\). The two potentials \(\dfrac{kq}{r}\) and \(-\dfrac{kq}{r}\) cancel, so \(V = 0\).

Field: the field of \(+q\) points away from \(+q\) and the field of \(-q\) points towards \(-q\). Their components along the line cancel, but their components parallel to the dipole axis both point from \(+q\) towards \(-q\) and add. So \(E \neq 0\).

1. \(E_r = -\dfrac{\partial}{\partial r}\left(\dfrac{kp\cos\theta}{r^2}\right) = \dfrac{2kp\cos\theta}{r^3}\).

   \(E_\theta = -\dfrac{1}{r}\dfrac{\partial}{\partial\theta}\left(\dfrac{kp\cos\theta}{r^2}\right) = \dfrac{kp\sin\theta}{r^3}\).

2. \(E = \sqrt{E_r^2 + E_\theta^2} = \dfrac{kp}{r^3}\sqrt{4\cos^2\theta + \sin^2\theta} = \dfrac{kp}{r^3}\sqrt{1 + 3\cos^2\theta}\), using \(\sin^2\theta = 1 - \cos^2\theta\).

   Also \(\tan\alpha = \dfrac{E_\theta}{E_r} = \dfrac{\sin\theta}{2\cos\theta} = \dfrac{1}{2}\tan\theta\), the result quoted in Question 7.

1. \(\tau = pE\sin\theta = 4 \times 10^{-9} \times 2 \times 10^5 \times \sin 30^\circ = 8 \times 10^{-4} \times \dfrac{1}{2} = 4 \times 10^{-4}\) N m.

2. \(\tau_{\max} = pE = 8 \times 10^{-4}\) N m, when \(\sin\theta = 1\), that is, \(\theta = 90^\circ\).

\(pE = 8 \times 10^{-4}\) J.

1. \(\theta = 0\): \(U = -pE = -8 \times 10^{-4}\) J.

2. \(\theta = 90^\circ\): \(U = 0\).

3. \(\theta = 180^\circ\): \(U = +pE = 8 \times 10^{-4}\) J.

The energy is lowest at \(\theta = 0\), with \(\vec{p}\) along \(\vec{E}\). That is the stable equilibrium. At \(180^\circ\) the torque is zero but the energy is highest: unstable.

\(W_{\text{ext}} = U_{\text{final}} - U_{\text{initial}} = pE(\cos\theta_1 - \cos\theta_2)\).

1. \(0^\circ \to 90^\circ\): \(W = pE(1 - 0) = 8 \times 10^{-4}\) J.

2. \(0^\circ \to 180^\circ\): \(W = pE(1 - (-1)) = 2pE = 1.6 \times 10^{-3}\) J.

1. \(F = qE = 2 \times 10^{-6} \times 2 \times 10^5 = 0.4\) N on each charge, along \(\vec{E}\) on \(+q\) and opposite to \(\vec{E}\) on \(-q\). The two forces are equal and opposite, so the net force is zero.

2. The force on \(+q\) pulls the upper end to the right and the force on \(-q\) pulls the lower end to the left. This pair turns the dipole clockwise, bringing \(\vec{p}\) towards \(\vec{E}\).

3. The two forces form a couple. The perpendicular distance between their lines is \(2a\sin\theta = 2 \times 10^{-3} \times \dfrac{1}{2} = 10^{-3}\) m.

   \(\tau = qE \times 2a\sin\theta = 0.4 \times 10^{-3} = 4 \times 10^{-4}\) N m.

   Check: \(pE\sin\theta = (q \cdot 2a)E\sin\theta = 4 \times 10^{-9} \times 2 \times 10^5 \times \dfrac{1}{2} = 4 \times 10^{-4}\) N m. Same.

\(pE = 4 \times 10^{-9} \times 2 \times 10^5 = 8 \times 10^{-4}\) J.

\(T = 2\pi\sqrt{\dfrac{8 \times 10^{-9}}{8 \times 10^{-4}}} = 2\pi\sqrt{10^{-5}} = 2\pi \times 3.16 \times 10^{-3} \approx 0.02\) s.

\(U = -pE\cos\theta\), so \(\dfrac{dU}{d\theta} = pE\sin\theta\).

The torque has magnitude \(pE\sin\theta\) and acts to decrease \(\theta\) (it turns \(\vec{p}\) towards \(\vec{E}\)). So \(\tau = -\dfrac{dU}{d\theta}\): the torque points towards lower energy, just as a force \(F = -\dfrac{dU}{dx}\) does.

\(\dfrac{dE}{dx} = 5 \times 10^6\) N/C per metre.

\(F = p\dfrac{dE}{dx} = 4 \times 10^{-9} \times 5 \times 10^6 = 0.02\) N.

The field grows along \(+x\) and \(\vec{p}\) is along \(\vec{E}\), so the dipole is pulled towards the stronger field: along \(+x\).

1. \(E = \dfrac{kQ}{x^2}\), so \(\dfrac{dE}{dx} = -\dfrac{2kQ}{x^3}\). The field weakens with distance from \(Q\).

2. \(|F| = p\left|\dfrac{dE}{dx}\right| = \dfrac{2kQp}{x^3} = \dfrac{2 \times 9 \times 10^9 \times 10^{-6} \times 4 \times 10^{-9}}{(0.1)^3} = \dfrac{7.2 \times 10^{-5}}{10^{-3}} = 0.072\) N.

   \(\vec{p}\) is along \(\vec{E}\), so the dipole is pulled towards the stronger field, that is, towards \(Q\). Check: \(-q\) is nearer \(Q\) than \(+q\), so the attraction wins.

1. Zero. The forces \(qE\) on \(+q\) and \(-q\) are equal and opposite. Only a torque acts.

2. Not zero. The charge in the stronger part of the field feels the larger force. With \(\vec{p}\) along \(\vec{E}\) the net force is towards the region of stronger field.

3. Not zero. With \(\vec{p}\) opposite to \(\vec{E}\) the net force is towards the region of weaker field.