The Workbook

1. Coulomb's law: \(F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2} = \dfrac{k q_1 q_2}{r^2}\) with \(k = 9 \times 10^9\) N m\(^2\)/C\(^2\).

   \(F = \dfrac{9 \times 10^9 \times 2 \times 10^{-6} \times 3 \times 10^{-6}}{(0.3)^2} = \dfrac{54 \times 10^{-3}}{0.09} = 0.6\) N.

2. Both charges are positive. Like charges repel, so the force is repulsive.

\(F = \dfrac{k|q_1 q_2|}{r^2}\), so \(r^2 = \dfrac{k|q_1 q_2|}{F} = \dfrac{9 \times 10^9 \times 4 \times 10^{-6} \times 6 \times 10^{-6}}{2.4} = \dfrac{0.216}{2.4} = 0.09\) m\(^2\).

\(r = 0.3\) m. The charges have opposite signs, so the force is attractive.

With \(q_1 = q_2 = q\): \(F = \dfrac{kq^2}{r^2}\), so \(q^2 = \dfrac{F r^2}{k} = \dfrac{3.6 \times (0.1)^2}{9 \times 10^9} = \dfrac{0.036}{9 \times 10^9} = 4 \times 10^{-12}\) C\(^2\).

\(q = 2 \times 10^{-6}\) C \(= 2\,\mu\)C.

1. In a medium the permittivity is \(\varepsilon = K\varepsilon_0\). So \(F = \dfrac{1}{4\pi K\varepsilon_0}\dfrac{q_1 q_2}{r^2}\). The force becomes \(\dfrac{1}{K}\) times the force in air.

2. \(F_{\text{liquid}} = \dfrac{F_{\text{air}}}{K} = \dfrac{0.8}{4} = 0.2\) N.

\(F \propto \dfrac{q_1 q_2}{r^2}\).

1. \(r \to 2r\): the force becomes \(\dfrac{F}{4}\).

2. \(q_1 \to 3q_1\): the force becomes \(3F\).

3. \(q_1 q_2 \to 4 q_1 q_2\) and \(r^2 \to \dfrac{r^2}{4}\): the force becomes \(4 \times 4 F = 16F\).

1. Magnitude: \(F = \dfrac{9 \times 10^9 \times 2 \times 10^{-6} \times 8 \times 10^{-6}}{(0.3)^2} = \dfrac{0.144}{0.09} = 1.6\) N.

   The charges are unlike, so \(q_2\) is pulled towards \(q_1\) at the origin, that is, along \(-x\). \(\vec{F}_{2} = -1.6\,\hat{i}\) N.

2. By Newton's third law the force on \(q_1\) is equal and opposite: \(\vec{F}_{1} = +1.6\,\hat{i}\) N. The charge \(q_1\) is pulled towards \(q_2\).

1. \(F = \dfrac{9 \times 10^9 \times 1 \times 10^{-6} \times 4 \times 10^{-6}}{(0.6)^2} = \dfrac{0.036}{0.36} = 0.1\) N on each ball.

2. The two forces are an action and reaction pair. They are equal in magnitude even though the charges and the masses are different.

3. \(a = \dfrac{F}{m}\). For the \(10\) g ball: \(a = \dfrac{0.1}{0.01} = 10\) m/s\(^2\). For the \(20\) g ball: \(a = \dfrac{0.1}{0.02} = 5\) m/s\(^2\).

   The forces are equal but the lighter ball accelerates more.

1. \(C\) is \(30\) cm from \(A\): \(F_{CA} = \dfrac{9 \times 10^9 \times 4 \times 10^{-6} \times 1 \times 10^{-6}}{(0.3)^2} = \dfrac{0.036}{0.09} = 0.4\) N. \(A\) and \(C\) are both positive, so this force pushes \(C\) away from \(A\).

   \(C\) is \(10\) cm from \(B\): \(F_{CB} = \dfrac{9 \times 10^9 \times 2 \times 10^{-6} \times 1 \times 10^{-6}}{(0.1)^2} = \dfrac{0.018}{0.01} = 1.8\) N. \(B\) is negative, so this force pulls \(C\) towards \(B\).

2. The two forces are opposite. Net force \(= 1.8 - 0.4 = 1.4\) N, towards \(B\).

3. Force on \(B\) due to \(A\): \(\dfrac{9 \times 10^9 \times 4 \times 10^{-6} \times 2 \times 10^{-6}}{(0.2)^2} = \dfrac{0.072}{0.04} = 1.8\) N, towards \(A\) (attraction).

   Force on \(B\) due to \(C\): \(1.8\) N, towards \(C\) (attraction). The two forces are equal and opposite, so the net force on \(B\) is zero.

1. Due to \(q_1\): \(F_1 = \dfrac{9 \times 10^9 \times 1 \times 10^{-6} \times 1 \times 10^{-6}}{(0.15)^2} = \dfrac{0.009}{0.0225} = 0.4\) N. Like charges, so \(q_0\) is pushed away from \(q_1\), along \(-x\).

   Due to \(q_2\): \(F_2 = \dfrac{9 \times 10^9 \times 1 \times 10^{-6} \times 3 \times 10^{-6}}{(0.3)^2} = \dfrac{0.027}{0.09} = 0.3\) N. Unlike charges, so \(q_0\) is pulled towards \(q_2\), along \(+y\).

2. The two forces are at right angles. \(F = \sqrt{0.4^2 + 0.3^2} = 0.5\) N.

   \(\tan\theta = \dfrac{0.3}{0.4} = \dfrac{3}{4}\), so \(\theta = 37^\circ\) measured from the \(-x\) direction towards \(+y\).

1. Both charges are positive, so the null point lies between them. Let it be \(x\) from the \(4\,\mu\)C charge and \((0.3 - x)\) from the \(1\,\mu\)C charge.

   Equal forces: \(\dfrac{k(4)q}{x^2} = \dfrac{k(1)q}{(0.3 - x)^2}\), so \(\dfrac{2}{x} = \dfrac{1}{0.3 - x}\).

   \(0.6 - 2x = x\), so \(x = 0.2\) m. The null point is \(20\) cm from the \(4\,\mu\)C charge.

2. No. The charge \(q\) cancels from the equation. The point is fixed by the two given charges only.

The charges are unlike, so between them the two forces on a third charge point the same way. The null point lies outside, on the side of the smaller charge.

Let the point be \(x\) from \(+4\,\mu\)C, so \((x - 0.3)\) from \(-1\,\mu\)C: \(\dfrac{4}{x^2} = \dfrac{1}{(x - 0.3)^2}\).

\(2(x - 0.3) = x\), so \(x = 0.6\) m. The point is \(60\) cm from \(+4\,\mu\)C, that is, \(30\) cm beyond the \(-1\,\mu\)C charge.

1. The two charges are equal, so the forces on a third charge cancel only at the midpoint, \(x = 20\) cm.

2. Now the charge at \(x = 0\) must also be in equilibrium. It is pushed by the other \(+4\,\mu\)C at \(40\) cm and pulled by \(q\) at \(20\) cm.

   \(\dfrac{k(4)(4)}{(0.4)^2} = \dfrac{k(4)|q|}{(0.2)^2}\), so \(|q| = 4 \times \dfrac{0.04}{0.16} = 1\,\mu\)C.

   The pull must balance the push, so \(q\) is negative: \(q = -1\,\mu\)C. By symmetry the charge at \(x = 40\) cm is then also in equilibrium.

1. \(F = \dfrac{kq^2}{a^2} = \dfrac{9 \times 10^9 \times (2 \times 10^{-6})^2}{(0.2)^2} = \dfrac{0.036}{0.04} = 0.9\) N from each of the other two charges.

2. The two forces have equal magnitude and the angle between them is \(60^\circ\).

   \(F_{\text{net}} = \sqrt{F^2 + F^2 + 2F^2\cos 60^\circ} = \sqrt{3}\,F = 0.9\sqrt{3} \approx 1.56\) N.

   By symmetry the net force lies along the bisector of the angle at that corner, pointing away from the centre of the triangle.

1. \(\dfrac{kq^2}{a^2} = \dfrac{9 \times 10^9 \times (10^{-6})^2}{(0.1)^2} = 0.9\) N.

   Due to \(B\) (distance \(a\)): \(0.9\) N along \(BA\). Due to \(D\) (distance \(a\)): \(0.9\) N along \(DA\).

   Due to \(C\) (distance \(a\sqrt{2}\)): \(\dfrac{kq^2}{2a^2} = 0.45\) N along \(CA\).

2. The forces due to \(B\) and \(D\) are at right angles. Their resultant is \(0.9\sqrt{2}\) N along the diagonal \(CA\), the same line as the force due to \(C\).

   \(F_{\text{net}} = 0.9\sqrt{2} + 0.45 = 0.9\left(\sqrt{2} + \dfrac{1}{2}\right) \approx 1.72\) N, along \(CA\) away from the square.

3. \(O\) is equidistant from the four equal charges. The forces due to \(A\) and \(C\) cancel, and so do the forces due to \(B\) and \(D\). The net force on \(Q\) is zero.

1. In a regular hexagon each corner is a distance \(a\) from the centre. The six forces on \(q_0\) are equal in size, and opposite corners act in opposite directions. The net force is zero.

2. The six forces add to zero, so the five remaining forces add to the negative of the missing force.

   The missing charge would push \(q_0\) away from its corner with force \(\dfrac{kqq_0}{a^2}\) (like charges). So the five charges give \(\dfrac{kqq_0}{a^2}\) directed towards the empty corner.

1. The five forces have equal size and are spaced at \(72^\circ\). By symmetry their vector sum is zero.

2. Replacing \(+q\) by \(-q\) at one corner reverses that one force. The change in the net force is \(-2\vec{F}_1\), where \(\vec{F}_1\) is the original force due to that charge.

   Originally \(\vec{F}_1\) pushed \(q_0\) away from the corner with magnitude \(\dfrac{kqq_0}{R^2}\). So the new net force is \(\dfrac{2kqq_0}{R^2}\), directed towards the \(-q\) charge.

1. Three forces act on each ball: the weight \(mg\) down, the tension \(T\) along the thread, and the electric force \(F\) horizontal.

   Vertical: \(T\cos\theta = mg\). Horizontal: \(T\sin\theta = F\). Dividing, \(F = mg\tan\theta = 0.12 \times 10 \times \dfrac{3}{4} = 0.9\) N.

2. \(F = \dfrac{kq^2}{r^2}\) with \(r = 0.3\) m: \(q^2 = \dfrac{0.9 \times 0.09}{9 \times 10^9} = 9 \times 10^{-12}\) C\(^2\).

   \(q = 3 \times 10^{-6}\) C \(= 3\,\mu\)C.

3. \(T = \dfrac{mg}{\cos\theta} = \dfrac{1.2}{4/5} = 1.5\) N.

1. \(F = \dfrac{9 \times 10^9 \times 6 \times 10^{-6} \times 2 \times 10^{-6}}{(0.3)^2} = \dfrac{0.108}{0.09} = 1.2\) N. Unlike charges, so attractive.

2. On touching, the total charge \(+6 - 2 = +4\,\mu\)C is shared equally between identical spheres: \(+2\,\mu\)C each.

   \(F = \dfrac{9 \times 10^9 \times (2 \times 10^{-6})^2}{(0.3)^2} = \dfrac{0.036}{0.09} = 0.4\) N. Like charges now, so repulsive.

1. Let the charges be \(q\) and \((Q - q)\) with \(Q = 8\,\mu\)C. \(F = \dfrac{k\,q(Q - q)}{r^2}\).

   The product \(q(Q - q)\) is largest when the two factors are equal, \(q = \dfrac{Q}{2}\). So the charge is divided into \(4\,\mu\)C and \(4\,\mu\)C.

2. \(F_{\max} = \dfrac{9 \times 10^9 \times (4 \times 10^{-6})^2}{(0.2)^2} = \dfrac{0.144}{0.04} = 3.6\) N.

1. \(F_e = \dfrac{ke^2}{r^2} = \dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{(5.3 \times 10^{-11})^2} = \dfrac{2.304 \times 10^{-28}}{2.81 \times 10^{-21}} \approx 8.2 \times 10^{-8}\) N.

2. \(F_g = \dfrac{G m_e m_p}{r^2} = \dfrac{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}}{2.81 \times 10^{-21}} \approx 3.6 \times 10^{-47}\) N.

   \(\dfrac{F_e}{F_g} = \dfrac{ke^2}{G m_e m_p} \approx \dfrac{8.2 \times 10^{-8}}{3.6 \times 10^{-47}} \approx 2.3 \times 10^{39}\). The distance cancels in the ratio.

1. \(n = \dfrac{|q|}{e} = \dfrac{3.2 \times 10^{-6}}{1.6 \times 10^{-19}} = 2 \times 10^{13}\) extra electrons.

2. \(\dfrac{2.4 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.5\). Charge comes in whole multiples of \(e\), so a charge of \(1.5e\) is not possible.

In the medium: \(F = \dfrac{kq_1 q_2}{K r^2}\). In air at distance \(r'\): \(F = \dfrac{kq_1 q_2}{r'^2}\).

Equal forces need \(r'^2 = K r^2\), so \(r' = r\sqrt{K} = 0.2 \times 2 = 0.4\) m \(= 40\) cm.

1. Distance from each charge to \(Q\): \(\sqrt{0.3^2 + 0.4^2} = 0.5\) m.

   \(F = \dfrac{9 \times 10^9 \times 5 \times 10^{-6} \times 5 \times 10^{-6}}{(0.5)^2} = \dfrac{0.225}{0.25} = 0.9\) N from each, directed away from that charge (repulsion).

2. Each force makes an angle \(\theta\) with the \(x\)-axis where \(\cos\theta = \dfrac{0.4}{0.5} = 0.8\). The \(y\)-components are equal and opposite and cancel.

   The \(x\)-components add: \(F_{\text{net}} = 2 \times 0.9 \times 0.8 = 1.44\) N along \(+x\), away from the origin.

3. At \(O\) the two charges are equidistant and on opposite sides. Their forces on \(Q\) are equal and opposite, so the net force is zero.

1. Each corner is a distance \(\dfrac{a}{\sqrt{2}}\) from the centre, so each charge acts on \(Q\) with magnitude \(\dfrac{kqQ}{(a/\sqrt{2})^2} = \dfrac{2kqQ}{a^2}\).

   \(A\) and \(C\) are opposite corners with equal charges \(+q\). Their pushes on \(Q\) are equal and opposite and cancel.

   \(B\) (\(+q\)) pushes \(Q\) away from \(B\), that is, towards \(D\). \(D\) (\(-q\)) pulls \(Q\) towards \(D\). These two add.

   \(F_{\text{net}} = 2 \times \dfrac{2kqQ}{a^2} = \dfrac{4kqQ}{a^2}\), directed from the centre towards \(D\).

2. \(F = \dfrac{4 \times 9 \times 10^9 \times 2 \times 10^{-6} \times 1 \times 10^{-6}}{(0.2)^2} = \dfrac{0.072}{0.04} = 1.8\) N.

1. \(E = \dfrac{F}{q} = \dfrac{0.01}{2 \times 10^{-6}} = 5000\) N/C. The field points along the force on the positive charge.

2. \(F = |q|E = 4 \times 10^{-6} \times 5000 = 0.02\) N. The charge is negative, so the force is opposite to \(E\).

\(E = \dfrac{kq}{r^2}\). At \(20\) cm: \(E = \dfrac{9 \times 10^9 \times 4 \times 10^{-6}}{(0.2)^2} = \dfrac{3.6 \times 10^4}{0.04} = 9 \times 10^5\) N/C.

At \(40\) cm the distance is doubled, so \(E\) is one quarter: \(2.25 \times 10^5\) N/C.

The charge is positive, so the field points away from the charge at every point.

1. Each charge is \(10\) cm from the midpoint: \(E_1 = E_2 = \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{(0.1)^2} = 1.8 \times 10^6\) N/C.

   The field of \(+2\,\mu\)C points away from it, towards \(-2\,\mu\)C. The field of \(-2\,\mu\)C points towards itself. Both fields point the same way.

   \(E = 1.8 \times 10^6 + 1.8 \times 10^6 = 3.6 \times 10^6\) N/C, directed from the positive charge to the negative charge.

2. For two equal positive charges the two fields at the midpoint are equal and opposite. \(E = 0\).

1. \(F = qE = 2 \times 10^{-6} \times 5000 = 0.01\) N. \(a = \dfrac{F}{m} = \dfrac{0.01}{0.001} = 10\) m/s\(^2\).

2. From rest, \(s = \dfrac{1}{2}at^2\): \(20 = \dfrac{1}{2} \times 10 \times t^2\), so \(t = 2\) s.

Distance from the origin: \(r = \sqrt{3^2 + 4^2} = 5\) m.

\(E = \dfrac{kq}{r^2} = \dfrac{9 \times 10^9 \times 5 \times 10^{-6}}{25} = 1800\) N/C, directed away from the origin along the line to the point.

The unit vector along this line is \(\left(\dfrac{3}{5}, \dfrac{4}{5}\right)\). So \(E_x = 1800 \times \dfrac{3}{5} = 1080\) N/C and \(E_y = 1800 \times \dfrac{4}{5} = 1440\) N/C.

1. \(\lambda = \dfrac{Q}{L} = \dfrac{10\,\mu\text{C}}{0.5\text{ m}} = 20\,\mu\)C/m.

2. \(q = \lambda \times 0.05 = 20 \times 0.05 = 1\,\mu\)C.

3. \(dq = \lambda\,dx\). Every formula for a continuous body starts by writing the charge of one small element this way.

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

2. \(E \propto \dfrac{1}{r}\), so the field halves when the distance doubles: \(r = 20\) cm.

3. For a long wire \(E \propto \dfrac{1}{r}\). For a point charge \(E \propto \dfrac{1}{r^2}\). The field of the wire falls more slowly with distance.

1. Each half of the rod is \(30\) cm long and \(P\) is \(40\) cm from the midpoint. The distance from \(P\) to either end is \(\sqrt{0.3^2 + 0.4^2} = 0.5\) m.

   By symmetry \(\alpha = \beta\) with \(\sin\alpha = \dfrac{0.3}{0.5} = 0.6\).

   \(E = \dfrac{k\lambda}{r}(2\sin\alpha) = \dfrac{9 \times 10^9 \times 5 \times 10^{-6}}{0.4} \times 1.2 = 2.25 \times 10^5 \times 0.6 = 1.35 \times 10^5\) N/C, perpendicular to the rod.

2. Total charge \(Q = \lambda L = 5 \times 0.6 = 3\,\mu\)C. As a point charge at \(40\) cm: \(E = \dfrac{9 \times 10^9 \times 3 \times 10^{-6}}{(0.4)^2} \approx 1.69 \times 10^5\) N/C.

   The point-charge value is larger. In the real rod the ends are farther from \(P\) than the midpoint, and their fields are tilted. Only the perpendicular components add.

1. The near end is at the foot of the perpendicular, so \(\alpha = 0\). The far end is at infinity, so \(\beta = 90^\circ\).

   \(E_\perp = \dfrac{k\lambda}{r}(\sin 0 + \sin 90^\circ) = \dfrac{k\lambda}{r} = \dfrac{9 \times 10^9 \times 2 \times 10^{-6}}{0.1} = 1.8 \times 10^5\) N/C, away from the rod.

2. \(E_\parallel = \dfrac{k\lambda}{r} = 1.8 \times 10^5\) N/C, directed away from the rod along its length (away from the far end).

   The two components are equal, so \(E = \sqrt{2} \times 1.8 \times 10^5 \approx 2.55 \times 10^5\) N/C, at \(45^\circ\) to the rod.

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

   Its field at \(P\) is \(dE = \dfrac{k\lambda\,dx}{x^2}\), and every element's field points the same way (away from the rod).

   \(E = k\lambda\displaystyle\int_a^{a+L}\dfrac{dx}{x^2} = k\lambda\left(\dfrac{1}{a} - \dfrac{1}{a + L}\right) = \dfrac{k\lambda L}{a(a + L)} = \dfrac{kQ}{a(a + L)}\).

2. \(E = \dfrac{9 \times 10^9 \times 6 \times 10^{-6}}{0.2 \times 0.5} = \dfrac{5.4 \times 10^4}{0.1} = 5.4 \times 10^5\) N/C.

3. For \(a \gg L\), \(a + L \approx a\) and \(E \approx \dfrac{kQ}{a^2}\). From far away the rod looks like a point charge \(Q\).

1. \(R^2 + x^2 = 0.09 + 0.16 = 0.25\), so \((R^2 + x^2)^{3/2} = (0.5)^3 = 0.125\) m\(^3\).

   \(E = \dfrac{9 \times 10^9 \times 5 \times 10^{-6} \times 0.4}{0.125} = \dfrac{1.8 \times 10^4}{0.125} = 1.44 \times 10^5\) N/C, along the axis away from the ring.

2. \(E = 0\) at the centre. Every element of the ring has a partner directly opposite it, at the same distance. Their fields at \(C\) are equal and opposite, so all cancel.

3. For \(x \gg R\), \(R^2 + x^2 \approx x^2\) and \(E \approx \dfrac{kQx}{x^3} = \dfrac{kQ}{x^2}\). From far away the ring looks like a point charge \(Q\).

1. \(x = \dfrac{R}{\sqrt{2}} = \dfrac{0.3}{1.414} \approx 0.21\) m.

2. Put \(x = \dfrac{R}{\sqrt{2}}\): \(R^2 + x^2 = \dfrac{3R^2}{2}\), so \((R^2 + x^2)^{3/2} = \dfrac{3\sqrt{3}}{2\sqrt{2}}R^3\).

   \(E_{\max} = \dfrac{kQ\,(R/\sqrt{2})}{(3\sqrt{3}/2\sqrt{2})R^3} = \dfrac{2kQ}{3\sqrt{3}\,R^2}\).

   \(E_{\max} = \dfrac{2 \times 9 \times 10^9 \times 5 \times 10^{-6}}{3 \times 1.732 \times 0.09} = \dfrac{9 \times 10^4}{0.4677} \approx 1.92 \times 10^5\) N/C.

\(\dfrac{2k\lambda}{R} = \dfrac{2 \times 9 \times 10^9 \times 2 \times 10^{-6}}{0.1} = 3.6 \times 10^5\) N/C.

1. \(\theta = \pi\): \(E = 3.6 \times 10^5 \sin 90^\circ = 3.6 \times 10^5\) N/C.

2. \(\theta = \dfrac{\pi}{2}\): \(E = 3.6 \times 10^5 \sin 45^\circ \approx 2.55 \times 10^5\) N/C.

3. \(\theta = 2\pi\): \(E = 3.6 \times 10^5 \sin 180^\circ = 0\). The full ring gives no field at its centre.

The length of the semicircle is \(\pi R\), so \(\lambda = \dfrac{Q}{\pi R}\).

\(E = \dfrac{2k\lambda}{R} = \dfrac{2kQ}{\pi R^2} = \dfrac{2 \times 9 \times 10^9 \times 2 \times 10^{-6}}{\pi \times 0.01} = \dfrac{3.6}{\pi} \times 10^6 \approx 1.15 \times 10^6\) N/C.

1. Each half is a semicircle, so each gives \(\dfrac{2k\lambda}{R} = \dfrac{2 \times 9 \times 10^9 \times 2 \times 10^{-6}}{0.1} = 3.6 \times 10^5\) N/C at \(O\).

   The positive upper half pushes a test charge at \(O\) away from itself, that is, downward. The negative lower half pulls the test charge towards itself, also downward.

2. The two fields point the same way. \(E = 2 \times 3.6 \times 10^5 = 7.2 \times 10^5\) N/C, directed from the positive half towards the negative half.

Let the arc run from \(\theta = -\dfrac{\pi}{2}\) to \(+\dfrac{\pi}{2}\), measured from the symmetry line through the centre.

An element of length \(R\,d\theta\) carries \(dq = \lambda R\,d\theta\) and gives \(dE = \dfrac{k\lambda R\,d\theta}{R^2} = \dfrac{k\lambda}{R}d\theta\) at the centre, directed away from the element.

The component along the symmetry line is \(dE\cos\theta\). The perpendicular components cancel in pairs by symmetry.

\(E = \dfrac{k\lambda}{R}\displaystyle\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta = \dfrac{k\lambda}{R}\left[\sin\theta\right]_{-\pi/2}^{\pi/2} = \dfrac{2k\lambda}{R}\).

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

2. \(\sqrt{x^2 + R^2} = \sqrt{0.16 + 0.09} = 0.5\) m, so \(\dfrac{x}{\sqrt{x^2 + R^2}} = \dfrac{0.4}{0.5} = 0.8\).

   \(E = 1000 \times (1 - 0.8) = 200\) N/C.

3. Very close to the surface \(x \to 0\), so the bracket becomes \(1\) and \(E = \dfrac{\sigma}{2\varepsilon_0} = 1000\) N/C.

1. \(E = \dfrac{\sigma}{2\varepsilon_0} = \dfrac{8.85 \times 10^{-9}}{2 \times 8.85 \times 10^{-12}} = 500\) N/C, perpendicular to the sheet, pointing away.

2. No. The formula has no distance in it. For an infinite sheet the field lines are parallel and evenly spaced, so the field is uniform.

3. Between the sheets, the field of \(+\sigma\) points away from it and the field of \(-\sigma\) points towards it. Both point the same way: \(E = 500 + 500 = 1000\) N/C, from \(+\sigma\) to \(-\sigma\).

   Outside, the two fields are opposite and equal: \(E = 0\).

Each sheet gives \(\dfrac{\sigma}{2\varepsilon_0} = 500\) N/C, pointing away from that sheet on both sides.

At \(P\) (left of both): both fields point left. \(E = 1000\) N/C to the left.

At \(Q\) (between): \(A\) pushes right, \(B\) pushes left. \(E = 0\).

At \(S\) (right of both): both fields point right. \(E = 1000\) N/C to the right.

1. \(E = \dfrac{\sigma}{2\varepsilon_0} = \dfrac{8.85 \times 10^{-8}}{2 \times 8.85 \times 10^{-12}} = 5000\) N/C, horizontal, away from the sheet.

   \(F = qE = 3 \times 10^{-6} \times 5000 = 0.015\) N, pushing the ball away from the sheet.

2. Weight \(mg = 0.002 \times 10 = 0.02\) N. Balance: \(T\sin\theta = qE\) and \(T\cos\theta = mg\).

   \(\tan\theta = \dfrac{qE}{mg} = \dfrac{0.015}{0.02} = \dfrac{3}{4}\), so \(\theta = 37^\circ\).

   \(T = \sqrt{(mg)^2 + (qE)^2} = \sqrt{0.02^2 + 0.015^2} = 0.025\) N.

1. Opposite sides are identical and equidistant from the centre. Their fields at the centre are equal and opposite. Both pairs cancel, so \(E = 0\).

2. The four sides gave zero, so the three remaining sides give the negative of the field of the missing side.

   One side is a rod of length \(a\) at perpendicular distance \(r = \dfrac{a}{2}\) from its midpoint. Each end subtends \(45^\circ\), so \(E_{\text{side}} = \dfrac{k\lambda}{a/2}(2\sin 45^\circ) = \dfrac{2\sqrt{2}\,k\lambda}{a}\).

   The missing side would push a test charge away from itself, towards the opposite side. So the three sides give \(\dfrac{2\sqrt{2}\,k\lambda}{a}\) directed towards the gap.