The Workbook

1. \(C = \dfrac{Q}{V} = \dfrac{4 \times 10^{-6}}{200} = 2 \times 10^{-8}\) F \(= 20\) nF.

2. \(Q = CV = 2 \times 10^{-8} \times 500 = 10^{-5}\) C \(= 10\,\mu\)C. The charge is proportional to the voltage; \(C\) is fixed by the geometry.

1. \(1\) F \(= 1\) C/V: one coulomb of charge for every volt of potential difference.

2. \(C = 4\pi\varepsilon_0 R\), so \(R = \dfrac{C}{4\pi\varepsilon_0} = kC = 9 \times 10^9 \times 1 = 9 \times 10^9\) m. This is more than a thousand times the radius of the Earth. A farad is a very large unit.

1. \(C = 4\pi\varepsilon_0 R = \dfrac{R}{k} = \dfrac{0.09}{9 \times 10^9} = 10^{-11}\) F \(= 10\) pF.

2. \(C = \dfrac{6.4 \times 10^6}{9 \times 10^9} \approx 7.1 \times 10^{-4}\) F \(\approx 711\,\mu\)F.

1. \(A = 100\) cm\(^2 = 10^{-2}\) m\(^2\), \(d = 10^{-3}\) m. \(C = \dfrac{\varepsilon_0 A}{d} = \dfrac{8.85 \times 10^{-12} \times 10^{-2}}{10^{-3}} = 8.85 \times 10^{-11}\) F \(= 88.5\) pF.

2. \(Q = CV = 8.85 \times 10^{-11} \times 12 \approx 1.06 \times 10^{-9}\) C \(= 1.06\) nC. \(E = \dfrac{V}{d} = \dfrac{12}{10^{-3}} = 1.2 \times 10^4\) V/m.

Each plate is a conductor with surface charge density \(\sigma = \dfrac{Q}{A}\) on its inner face. The field between the plates is \(E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0 A}\), uniform and perpendicular to the plates.

The potential difference is \(V = Ed = \dfrac{Qd}{\varepsilon_0 A}\).

\(C = \dfrac{Q}{V} = \dfrac{\varepsilon_0 A}{d}\). The charge cancels: \(C\) depends only on the geometry.

1. \(C \propto \dfrac{A}{d}\). Doubling \(d\) gives \(\dfrac{C}{2}\). Doubling \(A\) gives \(2C\).

2. \(A = \dfrac{Cd}{\varepsilon_0} = \dfrac{1 \times 10^{-3}}{8.85 \times 10^{-12}} \approx 1.13 \times 10^8\) m\(^2\), a square more than \(10\) km on a side. Real farad-sized capacitors need dielectrics and very thin gaps.

1. \(C = \dfrac{ab}{k(b - a)} = \dfrac{0.09 \times 0.1}{9 \times 10^9 \times 0.01} = \dfrac{0.009}{9 \times 10^7} = 10^{-10}\) F \(= 100\) pF.

   The isolated \(9\) cm sphere had \(10\) pF. Bringing the earthed shell close, \(1\) cm away, makes the capacitance \(10\) times larger.

2. For \(b \to \infty\), \(\dfrac{ab}{b - a} \to a\), so \(C \to 4\pi\varepsilon_0 a\). With the outer shell at infinity, the inner sphere is simply an isolated sphere.

\(C = \dfrac{2\pi\varepsilon_0 L}{\ln(b/a)} = \dfrac{2\pi \times 8.85 \times 10^{-12} \times 1}{1} \approx 5.56 \times 10^{-11}\) F \(= 55.6\) pF.

1. \(\sigma = \dfrac{Q}{A} = \dfrac{8.85 \times 10^{-9}}{10^{-2}} = 8.85 \times 10^{-7}\) C/m\(^2\).

2. Between: \(E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{8.85 \times 10^{-7}}{8.85 \times 10^{-12}} = 10^5\) V/m. Outside, the fields of the two plates cancel: \(E = 0\).

3. \(V = Ed = 10^5 \times 10^{-3} = 100\) V. Check: \(C = \dfrac{Q}{V} = \dfrac{8.85 \times 10^{-9}}{100} = 88.5\) pF, as in Question 4.

1. Near the edges of the plates the field lines bulge outward instead of running straight across. The field there is weaker and not uniform. This is fringing.

2. Fringing may be neglected when the separation is much smaller than the size of the plates. The bulging lines store some extra field outside the gap, so the true capacitance is slightly larger than \(\dfrac{\varepsilon_0 A}{d}\).

\(Q = CV = 2 \times 10^{-8} \times 200 = 4 \times 10^{-6}\) C.

\(\dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 2 \times 10^{-8} \times 200^2 = 4 \times 10^{-4}\) J.

\(\dfrac{1}{2}QV = \dfrac{1}{2} \times 4 \times 10^{-6} \times 200 = 4 \times 10^{-4}\) J. \(\dfrac{Q^2}{2C} = \dfrac{16 \times 10^{-12}}{4 \times 10^{-8}} = 4 \times 10^{-4}\) J. All three agree.

1. With the battery gone the charge has nowhere to go, so \(Q\) stays fixed. \(U = \dfrac{Q^2}{2C}\) doubles when \(C\) halves: \(U = 8 \times 10^{-4}\) J.

2. The plates attract each other. Pulling them apart needs an outside force, and the work done by that force, \(4 \times 10^{-4}\) J, is the extra energy.

The battery holds \(V\) at \(200\) V. New capacitance \(10\) nF, so \(Q = CV = 10^{-8} \times 200 = 2 \times 10^{-6}\) C \(= 2\,\mu\)C.

\(U = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 10^{-8} \times 4 \times 10^4 = 2 \times 10^{-4}\) J. Half the charge has flowed back into the battery.

1. \(u = \dfrac{1}{2}\varepsilon_0 E^2 = \dfrac{1}{2} \times 8.85 \times 10^{-12} \times 10^{10} \approx 0.044\) J/m\(^3\).

2. Volume \(= Ad = 10^{-2} \times 10^{-3} = 10^{-5}\) m\(^3\). \(U = 0.044 \times 10^{-5} = 4.4 \times 10^{-7}\) J.

   Check: \(\dfrac{1}{2}QV = \dfrac{1}{2} \times 8.85 \times 10^{-9} \times 100 = 4.4 \times 10^{-7}\) J. The energy of a capacitor lives in the field between its plates.

\(W = \displaystyle\int_0^Q \dfrac{q}{C}\,dq = \dfrac{1}{C}\left[\dfrac{q^2}{2}\right]_0^Q = \dfrac{Q^2}{2C}\).

This work is stored as \(U\). With \(Q = CV\): \(U = \dfrac{(CV)^2}{2C} = \dfrac{1}{2}CV^2\), and also \(U = \dfrac{1}{2}QV\).

The factor \(\dfrac{1}{2}\) appears because the voltage grows from \(0\) to \(V\) as the charge builds up; the average voltage during charging is \(\dfrac{V}{2}\).

1. \(Q = CV = 2 \times 10^{-8} \times 200 = 4\,\mu\)C flows through the battery. The battery moves this charge through its full \(200\) V: \(W = QV = 4 \times 10^{-6} \times 200 = 8 \times 10^{-4}\) J.

2. Stored: \(U = \dfrac{1}{2}QV = 4 \times 10^{-4}\) J, only half of the battery's work. The other half is lost as heat in the wires (and as radiation), whatever the resistance.

\(F = \dfrac{Q^2}{2\varepsilon_0 A} = \dfrac{(8.85 \times 10^{-9})^2}{2 \times 8.85 \times 10^{-12} \times 10^{-2}} = \dfrac{7.83 \times 10^{-17}}{1.77 \times 10^{-13}} \approx 4.4 \times 10^{-4}\) N.

Check: \(\dfrac{1}{2}QE = \dfrac{1}{2} \times 8.85 \times 10^{-9} \times 10^5 \approx 4.4 \times 10^{-4}\) N. Same.

A plate cannot push on itself. The force on a plate comes only from the field of the other plate, which is \(\dfrac{\sigma}{2\varepsilon_0} = \dfrac{E}{2}\).

The total field \(E\) between the plates is the sum of the two plates' fields, \(\dfrac{E}{2} + \dfrac{E}{2}\). So \(F = Q \times \dfrac{E}{2} = \dfrac{1}{2}QE\).

\(F = -\dfrac{dU}{dx} = -\dfrac{Q^2}{2\varepsilon_0 A}\).

The negative sign means the force acts to decrease \(x\): the plates pull towards each other. The force is attractive, with magnitude \(\dfrac{Q^2}{2\varepsilon_0 A}\), the same as in Question 17.

\(\dfrac{F}{A} = \dfrac{4.4 \times 10^{-4}}{10^{-2}} = 0.044\) N/m\(^2\).

\(\dfrac{\sigma^2}{2\varepsilon_0} = \dfrac{(8.85 \times 10^{-7})^2}{2 \times 8.85 \times 10^{-12}} = 0.044\) N/m\(^2\). They are the same thing: each plate is a charged conductor pulled by the field of the other, exactly the electrostatic pressure.

1. All the stored energy goes to the bulb: \(U = \dfrac{1}{2}CV^2 = 4 \times 10^{-4}\) J.

2. \(P = \dfrac{U}{t} = \dfrac{4 \times 10^{-4}}{2 \times 10^{-3}} = 0.2\) W.

Halving \(d\) doubles \(C\): \(C = 2 \times 88.5 = 177\) pF.

The battery keeps \(V = 12\) V, so \(Q = CV = 177 \times 10^{-12} \times 12 \approx 2.12\) nC, double the old charge.

\(E = \dfrac{V}{d} = \dfrac{12}{0.5 \times 10^{-3}} = 2.4 \times 10^4\) V/m, double the old field.

\(U = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 1.77 \times 10^{-10} \times 144 \approx 1.27 \times 10^{-8}\) J, double the old energy. The battery supplies the extra charge and energy.

1. Series: \(\dfrac{1}{C} = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}\), so \(C = \dfrac{6}{5} = 1.2\,\mu\)F. Less than the smaller one.

2. Parallel: \(C = 2 + 3 = 5\,\mu\)F. More than the larger one.

1. Series: \(C = \dfrac{6}{3} = 2\,\mu\)F.

2. Parallel: \(C = 3 \times 6 = 18\,\mu\)F.

3. Two in parallel give \(12\,\mu\)F. In series with \(6\,\mu\)F: \(\dfrac{1}{C} = \dfrac{1}{12} + \dfrac{1}{6} = \dfrac{3}{12}\), so \(C = 4\,\mu\)F.

1. \(C = 1.2\,\mu\)F, so \(Q = CV = 1.2 \times 100 = 120\,\mu\)C. In series the same charge sits on every capacitor.

2. \(V_1 = \dfrac{Q}{C_1} = \dfrac{120}{2} = 60\) V and \(V_2 = \dfrac{120}{3} = 40\) V. Check: \(60 + 40 = 100\) V. The smaller capacitor takes the larger share of the voltage.

In parallel both have the full \(100\) V. \(Q_1 = 2 \times 100 = 200\,\mu\)C and \(Q_2 = 3 \times 100 = 300\,\mu\)C.

Total \(= 500\,\mu\)C, the same as \(C_{\text{eq}}V = 5 \times 100\).

The plate of \(C_1\) joined to the plate of \(C_2\) forms an isolated conductor with no net charge to begin with. When \(+Q\) appears on the outer plate of \(C_1\), it pulls \(-Q\) onto the inner plate. That leaves \(+Q\) on the joined plate of \(C_2\). So both carry \(Q\).

The voltages add: \(V = V_1 + V_2 = \dfrac{Q}{C_1} + \dfrac{Q}{C_2}\).

The combination has \(C = \dfrac{Q}{V}\), so \(\dfrac{1}{C} = \dfrac{V}{Q} = \dfrac{1}{C_1} + \dfrac{1}{C_2}\).

Series: \(C_s = \dfrac{C}{n}\). Parallel: \(C_p = nC\).

\(\dfrac{C_p}{C_s} = \dfrac{nC}{C/n} = n^2\). For \(n = 4\) the ratio is \(16\).

1. The two \(3\,\mu\)F in parallel give \(6\,\mu\)F. In series with the other \(6\,\mu\)F: \(C = \dfrac{6 \times 6}{6 + 6} = 3\,\mu\)F.

2. \(Q = CV = 3 \times 12 = 36\,\mu\)C.

3. The \(6\,\mu\)F carries the full \(36\,\mu\)C, so \(V = \dfrac{36}{6} = 6\) V. The parallel pair gets the remaining \(6\) V.

   Each \(3\,\mu\)F has \(6\) V across it: \(Q = 3 \times 6 = 18\,\mu\)C. Check: \(18 + 18 = 36\,\mu\)C.

1. Balance condition: \(\dfrac{C_{AP}}{C_{PB}} = \dfrac{C_{AQ}}{C_{QB}}\). Here \(\dfrac{2}{4} = \dfrac{1}{2}\) and \(\dfrac{3}{6} = \dfrac{1}{2}\). Both series arms divide the voltage in the same ratio, so \(V_P = V_Q\).

2. No potential difference across the \(5\,\mu\)F, so its charge is \(0\). It can be removed without changing anything.

3. Upper arm: \(\dfrac{2 \times 4}{6} = \dfrac{4}{3}\,\mu\)F. Lower arm: \(\dfrac{3 \times 6}{9} = 2\,\mu\)F. In parallel: \(C_{AB} = \dfrac{4}{3} + 2 = \dfrac{10}{3} \approx 3.33\,\mu\)F.

\(\dfrac{1}{C} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3 + 2 + 1}{6} = 1\), so \(C = 1\,\mu\)F.

\(Q = CV = 1 \times 60 = 60\,\mu\)C on each capacitor.

\(V_1 = \dfrac{60}{2} = 30\) V, \(V_2 = \dfrac{60}{3} = 20\) V, \(V_3 = \dfrac{60}{6} = 10\) V. Check: \(30 + 20 + 10 = 60\) V.

1. \(C = \dfrac{4 \times 6}{10} = 2.4\,\mu\)F. \(Q = 2.4 \times 100 = 240\,\mu\)C.

2. The wire short-circuits the \(6\,\mu\)F: its two plates are now at the same potential and it holds no charge. The full \(100\) V falls across the \(4\,\mu\)F: \(Q = 4 \times 100 = 400\,\mu\)C.

1. Series pair: \(\dfrac{2 \times 3}{5} = 1.2\,\mu\)F. In parallel with \(6\,\mu\)F: \(C = 7.2\,\mu\)F.

2. The \(6\,\mu\)F has \(20\) V across it: \(Q = 6 \times 20 = 120\,\mu\)C.

   The series pair also has \(20\) V across it: \(Q = 1.2 \times 20 = 24\,\mu\)C, the same on the \(2\,\mu\)F and on the \(3\,\mu\)F.

3. \(V_2 = \dfrac{24}{2} = 12\) V and \(V_3 = \dfrac{24}{3} = 8\) V. Check: \(12 + 8 = 20\) V.

\(U = \dfrac{1}{2}C_{\text{eq}}V^2\).

1. Series: \(U = \dfrac{1}{2} \times 1.2 \times 10^{-6} \times 100^2 = 6 \times 10^{-3}\) J.

2. Parallel: \(U = \dfrac{1}{2} \times 5 \times 10^{-6} \times 10^4 = 0.025\) J. The parallel combination stores more because it holds more charge at the same voltage.

1. Charge before: \(Q = 2 \times 100 = 200\,\mu\)C on the \(2\,\mu\)F, none on the \(3\,\mu\)F. Charge is conserved, so the total stays \(200\,\mu\)C.

   After the switch closes the two are in parallel: \(V = \dfrac{Q_{\text{total}}}{C_1 + C_2} = \dfrac{200}{5} = 40\) V.

2. \(Q_1 = 2 \times 40 = 80\,\mu\)C and \(Q_2 = 3 \times 40 = 120\,\mu\)C. Check: \(80 + 120 = 200\,\mu\)C.

Before: \(U_i = \dfrac{1}{2} \times 2 \times 10^{-6} \times 100^2 = 0.01\) J.

After: \(U_f = \dfrac{1}{2} \times 5 \times 10^{-6} \times 40^2 = 4 \times 10^{-3}\) J.

Heat \(= U_i - U_f = 6 \times 10^{-3}\) J. Check with the formula of Question 17: \(\dfrac{2 \times 3 \times 10^{-12} \times 100^2}{2 \times 5 \times 10^{-6}} = 6 \times 10^{-3}\) J.

1. Total charge \(= 2 \times 100 + 3 \times 50 = 350\,\mu\)C. \(V = \dfrac{350}{5} = 70\) V.

2. \(H = \dfrac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)} = \dfrac{2 \times 3 \times 10^{-12} \times 50^2}{2 \times 5 \times 10^{-6}} = \dfrac{1.5 \times 10^{-8}}{10^{-5}} = 1.5 \times 10^{-3}\) J.

1. Positive to negative: the charges partly cancel. Net charge \(= 200 - 150 = 50\,\mu\)C. \(V = \dfrac{50}{5} = 10\) V.

2. Before: \(U_i = \dfrac{1}{2} \times 2 \times 10^{-6} \times 10^4 + \dfrac{1}{2} \times 3 \times 10^{-6} \times 2500 = 0.01 + 3.75 \times 10^{-3} = 0.01375\) J.

   After: \(U_f = \dfrac{1}{2} \times 5 \times 10^{-6} \times 10^2 = 2.5 \times 10^{-4}\) J. Heat \(= 0.01375 - 0.00025 = 0.0135\) J. Almost all the energy is lost.

Total charge \(Q = C_1 V_1 + C_2 V_2\) is conserved. Final common voltage: \(V = \dfrac{C_1 V_1 + C_2 V_2}{C_1 + C_2}\).

\(U_i = \dfrac{1}{2}C_1 V_1^2 + \dfrac{1}{2}C_2 V_2^2\) and \(U_f = \dfrac{1}{2}(C_1 + C_2)V^2 = \dfrac{(C_1 V_1 + C_2 V_2)^2}{2(C_1 + C_2)}\).

\(H = U_i - U_f = \dfrac{(C_1 V_1^2 + C_2 V_2^2)(C_1 + C_2) - (C_1 V_1 + C_2 V_2)^2}{2(C_1 + C_2)}\).

Expanding the numerator, the \(C_1^2 V_1^2\) and \(C_2^2 V_2^2\) terms cancel, leaving \(C_1 C_2 (V_1^2 + V_2^2 - 2V_1 V_2) = C_1 C_2 (V_1 - V_2)^2\). So \(H = \dfrac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}\).

1. \(U_i = \dfrac{1}{2}CV^2\). The charge \(CV\) spreads over \(2C\), so the common voltage is \(\dfrac{V}{2}\) and \(U_f = \dfrac{1}{2}(2C)\left(\dfrac{V}{2}\right)^2 = \dfrac{1}{4}CV^2\). Half the energy is lost.

2. It becomes heat in the wires and, if the wires are very good, electromagnetic radiation from the spark. The fraction lost is fixed by charge conservation and does not depend on the resistance. Resistance only changes how fast the loss happens.

Charge \(= 4 \times 200 = 800\,\mu\)C, shared equally by two identical capacitors: \(400\,\mu\)C each. \(V = \dfrac{400}{4} = 100\) V.

\(U_i = \dfrac{1}{2} \times 4 \times 10^{-6} \times 200^2 = 0.08\) J. \(U_f = \dfrac{1}{2} \times 8 \times 10^{-6} \times 100^2 = 0.04\) J. Heat \(= 0.04\) J, half of the original energy.

In series both carry the same charge \(Q\). Limits: \(Q \le 2 \times 300 = 600\,\mu\)C from the first, and \(Q \le 3 \times 150 = 450\,\mu\)C from the second.

The smaller limit wins: \(Q_{\max} = 450\,\mu\)C. Then \(V_1 = \dfrac{450}{2} = 225\) V (safe) and \(V_2 = 150\) V.

Largest safe total: \(225 + 150 = 375\) V.

The bridge is balanced, since \(\dfrac{C}{C} = \dfrac{C}{C}\). \(P\) and \(Q\) are at the same potential, so the capacitor \(PQ\) carries no charge.

Remove it: each arm is two \(C\) in series, \(\dfrac{C}{2}\). The two arms in parallel give \(\dfrac{C}{2} + \dfrac{C}{2} = C\).

1. From Question 3 each holds \(120\,\mu\)C. In parallel the total is \(240\,\mu\)C over \(5\,\mu\)F: \(V = \dfrac{240}{5} = 48\) V.

2. Before: \(U_i = \dfrac{Q^2}{2C_{\text{eq}}} = \dfrac{(120 \times 10^{-6})^2}{2 \times 1.2 \times 10^{-6}} = 6 \times 10^{-3}\) J.

   After: \(U_f = \dfrac{1}{2} \times 5 \times 10^{-6} \times 48^2 = 5.76 \times 10^{-3}\) J. Heat \(= 2.4 \times 10^{-4}\) J. Check: \(V_1 = 60\) V and \(V_2 = 40\) V in the formula of Question 17 give the same.

1. \(C = KC_0 = 4 \times 88.5 = 354\) pF.

2. With \(Q\) fixed, \(V = \dfrac{Q}{C}\) falls to \(\dfrac{1}{4}\), and \(E = \dfrac{V}{d}\) falls to \(\dfrac{E_0}{4}\). The dielectric cancels three quarters of the field.

Battery disconnected: \(Q\) is fixed at \(CV = 2 \times 10^{-8} \times 200 = 4\,\mu\)C.

\(C = 4 \times 20 = 80\) nF. \(V = \dfrac{Q}{C} = \dfrac{4 \times 10^{-6}}{8 \times 10^{-8}} = 50\) V, a quarter of the old value. \(E = \dfrac{V}{d}\) also falls to \(\dfrac{E_0}{4}\).

\(U = \dfrac{Q^2}{2C}\) falls to a quarter: \(U = \dfrac{1}{2} \times 8 \times 10^{-8} \times 50^2 = 10^{-4}\) J (it was \(4 \times 10^{-4}\) J).

Battery connected: \(V\) is fixed at \(200\) V. \(C = 80\) nF as before.

\(Q = CV = 8 \times 10^{-8} \times 200 = 16\,\mu\)C, four times the old charge. \(E = \dfrac{V}{d}\) is unchanged.

\(U = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 8 \times 10^{-8} \times 4 \times 10^4 = 1.6 \times 10^{-3}\) J, four times the old energy.

1. The field pulls the slab into the gap. The energy lost by the capacitor is the work done by this force on the slab. It appears as kinetic energy of the slab, or as work against whoever holds it.

2. The battery pushes the extra charge \(12\,\mu\)C through \(200\) V. It supplies \(2.4 \times 10^{-3}\) J, of which \(1.2 \times 10^{-3}\) J is stored and the rest is work done on the slab.

1. The field of the plates polarises the dielectric: its molecules turn or stretch. A negative layer appears on the face near the positive plate, and a positive layer near the negative plate. These induced layers produce a field opposite to that of the plates, so the net field inside is smaller.

2. \(\sigma_i = 8.85 \times 10^{-7}\left(1 - \dfrac{1}{4}\right) = 6.64 \times 10^{-7}\) C/m\(^2\). The net density \(\sigma - \sigma_i = \dfrac{\sigma}{K}\) produces the reduced field \(\dfrac{E_0}{K}\).

\(d - t + \dfrac{t}{K} = 1 - 0.5 + \dfrac{0.5}{4} = 0.625\) mm. The gap behaves like an air gap of \(0.625\) mm.

\(C = \dfrac{\varepsilon_0 A}{0.625\text{ mm}} = 88.5 \times \dfrac{1}{0.625} \approx 142\) pF.

A metal slab has no field inside it; it acts like \(K \to \infty\), so \(\dfrac{t}{K} \to 0\). \(C = \dfrac{\varepsilon_0 A}{d - t} = \dfrac{88.5}{0.5} = 177\) pF.

No. Only the total thickness of air, \(d - t\), matters, wherever the slab sits.

Air part: thickness \((d - t)\), \(C_1 = \dfrac{\varepsilon_0 A}{d - t}\). Dielectric part: thickness \(t\), \(C_2 = \dfrac{K\varepsilon_0 A}{t}\). They carry the same charge, so they are in series.

\(\dfrac{1}{C} = \dfrac{d - t}{\varepsilon_0 A} + \dfrac{t}{K\varepsilon_0 A} = \dfrac{d - t + t/K}{\varepsilon_0 A}\), so \(C = \dfrac{\varepsilon_0 A}{d - t + t/K}\).

The two halves have the same voltage across them, so they are in parallel. Each has area \(\dfrac{A}{2}\).

\(C = \dfrac{K_1\varepsilon_0 (A/2)}{d} + \dfrac{K_2\varepsilon_0 (A/2)}{d} = \dfrac{K_1 + K_2}{2}C_0 = \dfrac{2 + 4}{2} \times 88.5 = 265.5\) pF.

The two layers carry the same charge, so they are in series. Each has thickness \(\dfrac{d}{2}\): \(C_1 = \dfrac{2K_1\varepsilon_0 A}{d}\) and \(C_2 = \dfrac{2K_2\varepsilon_0 A}{d}\).

\(C = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{2\varepsilon_0 A}{d}\cdot\dfrac{K_1 K_2}{K_1 + K_2} = 2 \times 88.5 \times \dfrac{8}{6} = 236\) pF.

1. \(C = 2 \times 88.5 = 177\) pF.

2. Half area with \(K = 2\) in parallel with half area of air: \(C = \dfrac{2 + 1}{2} \times 88.5 = 132.75\) pF. Filling half the area gives less than filling all of it, as expected.

1. \(Q = CV = 141.6 \times 10^{-12} \times 12 \approx 1.7 \times 10^{-9}\) C \(= 1.7\) nC.

2. In the air part \(E_0 = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0 A} = \dfrac{1.7 \times 10^{-9}}{8.85 \times 10^{-12} \times 10^{-2}} \approx 1.92 \times 10^4\) V/m.

   In the dielectric \(E = \dfrac{E_0}{K} = 4.8 \times 10^3\) V/m.

   Check: \(V = 1.92 \times 10^4 \times 0.5 \times 10^{-3} + 4.8 \times 10^3 \times 0.5 \times 10^{-3} = 9.6 + 2.4 = 12\) V.

Air: \(V_{\max} = E_{\max}d = 3 \times 10^6 \times 10^{-3} = 3000\) V. Dielectric: \(V_{\max} = 12 \times 10^6 \times 10^{-3} = 12000\) V.

\(U_{\max} = \dfrac{1}{2}CV_{\max}^2\). \(C\) rises \(4\) times and \(V_{\max}\) rises \(4\) times, so \(U_{\max}\) rises \(4 \times 16 = 64\) times. This is why real capacitors use dielectrics.

1. Every slice carries the same charge \(Q\) on its two faces, and the voltages of the slices add. The slices are in series.

2. For series capacitors the reciprocals add: \(\dfrac{1}{C} = \displaystyle\int\dfrac{dx}{K(x)\varepsilon_0 A} = \dfrac{1}{\varepsilon_0 A}\int_0^d\dfrac{dx}{K(x)}\).

   With \(K = 2 + \dfrac{2x}{d}\): \(\displaystyle\int_0^d\dfrac{dx}{2 + 2x/d} = \dfrac{d}{2}\left[\ln\left(2 + \dfrac{2x}{d}\right)\right]_0^d = \dfrac{d}{2}\ln\dfrac{4}{2} = \dfrac{d\ln 2}{2}\).

   So \(\dfrac{1}{C} = \dfrac{d\ln 2}{2\varepsilon_0 A}\) and \(C = \dfrac{2\varepsilon_0 A}{d\ln 2}\).

3. \(C = \dfrac{2}{\ln 2}C_0 = \dfrac{2 \times 88.5}{0.693} \approx 255\) pF. This lies between \(2C_0\) and \(4C_0\), as it should.

A strip of width \(dy\) at distance \(y\) along the plate (length \(L\), breadth \(b\)) has \(dC = \dfrac{K(y)\varepsilon_0\,b\,dy}{d}\). All strips have the same voltage, so they are in parallel and their capacitances add.

\(C = \dfrac{\varepsilon_0 b}{d}\displaystyle\int_0^L K(y)\,dy\). For a linear \(K\) the integral is \(L\) times the average of \(K\): \(\dfrac{K_1 + K_2}{2}L\).

\(C = \dfrac{(K_1 + K_2)\varepsilon_0 A}{2d} = \dfrac{2 + 4}{2} \times 88.5 = 265.5\) pF, the same as the side-by-side case.

The two layers are in series: \(\dfrac{1}{C} = \dfrac{d/2}{K_1\varepsilon_0 A} + \dfrac{d/2}{K_2\varepsilon_0 A} = \dfrac{d}{2\varepsilon_0 A}\left(\dfrac{1}{K_1} + \dfrac{1}{K_2}\right) = \dfrac{d}{2\varepsilon_0 A}\cdot\dfrac{K_1 + K_2}{K_1 K_2}\).

Inverting: \(C = \dfrac{2\varepsilon_0 A}{d}\cdot\dfrac{K_1 K_2}{K_1 + K_2}\).

\(C = KC_0 = 4 \times 100 = 400\) pF. Filling the whole field region with a dielectric divides the field by \(K\) for the same charge. So the voltage also falls by \(K\), whatever the shape of the capacitor.

1. Pulled in. With \(Q\) fixed, \(U = \dfrac{Q^2}{2C}\), and \(C\) grows as more of the slab enters. The system moves towards lower energy, so the force on the slab points into the gap.

2. No. With the battery connected the capacitor's own energy rises, but the battery supplies twice that amount. The total energy of capacitor plus battery still falls as the slab enters. The slab is pulled in either way.

\(W_{\text{agent}} = U_{\text{final}} - U_{\text{initial}} = 10^{-4} - 4 \times 10^{-4} = -3 \times 10^{-4}\) J.

Negative work: the field pulls the slab in, and the agent must hold it back to insert it slowly.

Extra charge \(\Delta Q = 16 - 4 = 12\,\mu\)C passes through the battery at \(200\) V: \(W_{\text{battery}} = 12 \times 10^{-6} \times 200 = 2.4 \times 10^{-3}\) J.

\(\Delta U = 1.6 \times 10^{-3} - 4 \times 10^{-4} = 1.2 \times 10^{-3}\) J.

Energy balance: \(W_{\text{battery}} + W_{\text{agent}} = \Delta U\), so \(W_{\text{agent}} = 1.2 \times 10^{-3} - 2.4 \times 10^{-3} = -1.2 \times 10^{-3}\) J. Again the slab is pulled in.

The filled capacitor has \(4 \times 20 = 80\) nF. Series: \(C = \dfrac{20 \times 80}{100} = 16\) nF. \(Q = CV = 16 \times 10^{-9} \times 100 = 1.6\,\mu\)C on each.

Air: \(V = \dfrac{1.6 \times 10^{-6}}{2 \times 10^{-8}} = 80\) V. Dielectric: \(V = \dfrac{1.6 \times 10^{-6}}{8 \times 10^{-8}} = 20\) V. Most of the voltage falls across the air capacitor.

1. The filled part (length \(x\)) and the empty part (length \(L - x\)) share the same voltage, so they are in parallel. With plate breadth \(b\), \(A = bL\).

   \(C = \dfrac{K\varepsilon_0 bx}{d} + \dfrac{\varepsilon_0 b(L - x)}{d} = \dfrac{\varepsilon_0 bL}{d}\left[1 + (K - 1)\dfrac{x}{L}\right] = C_0\left[1 + (K - 1)\dfrac{x}{L}\right]\).

2. \(x = \dfrac{L}{4}\): \(C = 88.5\left[1 + 3 \times \dfrac{1}{4}\right] = 88.5 \times 1.75 \approx 155\) pF.