The last topic of this lesson, a dielectric constant that varies from point to point, goes beyond the Class 12 syllabus. It is JEE material and needs one integral. The rest of the lesson, slabs that fill part of the gap and dielectrics side by side or stacked, belongs to every syllabus.
A slab thinner than the gap
Slide a slab of thickness \(t\) and constant \(K\) into a gap \(d\), parallel to the plates. It fills the whole area but only part of the gap. Imagine a thin metal foil pressed onto each face of the slab. The field meets the foil at right angles, so the foil changes nothing. But now the gap is plainly a chain of capacitors: an air gap, the slab, and another air gap.
The same free charge \(Q\) passes from plate to foil to plate, so the parts are in series. The two air parts add up to a thickness \(d - t\), wherever the slab sits. Each part has the area \(A\). So \(1/C\) \(= (d - t)/\varepsilon_0A + t/K\varepsilon_0A\).
The slab acts like air of thickness \(t/K\). A slab of \(K = 4\) and \(t = 0.5\,\text{mm}\) in the \(1\,\text{mm}\) gap of Lesson 1 makes the gap act like \(1 - 0.5 + 0.125\) \(= 0.625\,\text{mm}\) of air. So \(C\) \(= 88.5 \times 1/0.625\) \(\approx 142\,\text{pF}\). A metal slab of the same thickness has no field inside at all, the case \(K \to \infty\). The gap then acts like \(0.5\,\text{mm}\) of air, and \(C = 177\,\text{pF}\), wherever the slab sits.
The capacitor with the \(0.5\,\text{mm}\) slab of \(K = 4\) is joined to \(12\,\text{V}\). Find the charge, the field in the air and in the slab, and check that the potential drops add to \(12\,\text{V}\). Take the \(\varepsilon_0\) route.
\(C \approx 141.6\,\text{pF}\), \(A = 10^{-2}\,\text{m}^2\), air thickness \(0.5\,\text{mm}\), slab \(0.5\,\text{mm}\).
\(Q = CV\); in the air \(E_0 = Q/\varepsilon_0A\); in the slab \(E = E_0/K\).
\(Q\) \(= 141.6 \times 10^{-12} \times 12\) \(\approx 1.7\,\text{nC}\). \(E_0\) \(= 1.7 \times 10^{-9}/(8.85 \times 10^{-12} \times 10^{-2})\) \(\approx 1.92 \times 10^4\,\text{V/m}\), and in the slab \(4.8 \times 10^3\,\text{V/m}\).
The plates hold \(1.7\,\text{nC}\). The air carries four times the field of the slab, and the drops \(9.6\,\text{V}\) and \(2.4\,\text{V}\) add to \(12\,\text{V}\).
A slab of \(K = 2\), half as thick as the gap, is placed in an \(88.5\,\text{pF}\) air capacitor. What is the new capacitance, in picofarads?
A metal slab, thinner than the gap and parallel to the plates, is moved from touching one plate to the middle of the gap. What happens to the capacitance?
Side by side and stacked
Fill the gap with two dielectrics, \(K_1\) and \(K_2\). There are two ways, shown in Fig 6.2. Side by side, each fills half the area across the whole gap. Stacked, each fills half the gap across the whole area. The question in each case is the one of Lesson 3: what do the two parts share?
Side by side, both parts run from one plate to the other, so they share the voltage \(V\). They are in parallel, each with half the area: \(C\) \(= K_1\varepsilon_0(A/2)/d + K_2\varepsilon_0(A/2)/d\) \(= \tfrac12(K_1 + K_2)C_0\). For the \(88.5\,\text{pF}\) capacitor with \(K_1 = 2\) and \(K_2 = 4\), \(C\) \(= 3 \times 88.5\) \(= 265.5\,\text{pF}\).
Stacked, the same charge passes through both layers, so they are in series. Each has the whole area and half the gap, so each is \(2K\varepsilon_0A/d\) \(= 2KC_0\). Then \(C\) \(= (2K_1C_0)(2K_2C_0)/(2K_1C_0 + 2K_2C_0)\) \(= 2C_0K_1K_2/(K_1 + K_2)\). With the same values, \(C\) \(= 2 \times 88.5 \times 8/6\) \(= 236\,\text{pF}\), less than side by side.
Half a gap filled side by side is a special case. Fill half the area with \(K = 2\) and leave the rest air: \(C\) \(= \tfrac12(2 + 1) \times 88.5\) \(= 132.75\,\text{pF}\). Fill the whole gap with \(K = 2\) and \(C = 177\,\text{pF}\). The half-filled capacitor has less, as it should.
The gap of the \(88.5\,\text{pF}\) air capacitor is filled with two stacked layers of equal thickness, \(K_1 = 3\) and \(K_2 = 6\). What is the capacitance, in picofarads?
Why are two dielectrics side by side, each running from plate to plate, treated as capacitors in parallel?
A slab sliding in
Now push a slab as thick as the gap in from the side, a distance \(x\) along plates of length \(L\). The part over the slab and the empty part both run from plate to plate, so they share the voltage. They are side by side, in parallel. With plates of breadth \(b\), \(C\) \(= K\varepsilon_0bx/d + \varepsilon_0b(L - x)/d\) \(= C_0[1 + (K - 1)x/L]\).
Move the slider. At a quarter of the way in, \(C\) \(= 88.5(1 + 3/4)\) \(\approx 155\,\text{pF}\); fully in, \(354\,\text{pF}\). With the battery removed, the charge \(1.06\,\text{nC}\) moves over onto the filled part, and the voltage falls. With the battery connected, the voltage stays \(12\,\text{V}\) and the battery feeds in more charge. In both chips the arrow \(F\) points into the gap.
Why is the slab always pulled in? With the battery removed, \(U = Q^2/2C\) falls as \(C\) grows, so the force \(-dU/dx\) points into the gap. With the battery connected, a rise \(dC\) raises the store by \(\tfrac12V^2\,dC\). But the battery supplies \(V\,dQ\) \(= V^2\,dC\), twice as much. The other \(\tfrac12V^2\,dC\) is work done on the slab, so again it is pulled in.
The pull comes from the fringing field at the edge of the plates, acting on the slab's induced charges. At fixed voltage it is \(F = \tfrac12V^2\,dC/dx\) \(= \tfrac12V^2C_0(K - 1)/L\). For our capacitor, with \(L = 10\,\text{cm}\) and \(12\,\text{V}\), \(F\) \(= \tfrac12 \times 144 \times 88.5 \times 10^{-12} \times 3/0.1\) \(\approx 1.9 \times 10^{-7}\,\text{N}\). It is tiny here, but at thousands of volts it can lift a liquid between the plates.
A slab of dielectric is half inside a charged capacitor that stays connected to its battery. The slab is let go. What does it do?
A capacitor of \(100\,\text{pF}\) in air has a slab of \(K = 3\), as thick as the gap, pushed half-way in along its length. What is the capacitance, in picofarads?
A dielectric constant that varies
Suppose \(K\) changes smoothly across the gap, from \(K = 2\) at one plate to \(K = 4\) at the other: \(K(x) = 2 + 2x/d\). Cut the gap into thin slices of thickness \(dx\), parallel to the plates. Each slice is a capacitor of area \(A\) and gap \(dx\), with capacitance \(K(x)\varepsilon_0A/dx\). Every slice carries the same charge, so the slices are in series, and their reciprocals add.
\[ \dfrac{1}{C} = \displaystyle\int_0^d \dfrac{dx}{K(x)\,\varepsilon_0 A} \]
Why. slices in series: add the reciprocals \(dx/K\varepsilon_0A\) across the gap.
?
Your turn. the integral of \(1/(2 + 2x/d)\) is \(\tfrac d2\ln(2 + 2x/d)\); put in the limits. Which line is it?
\[ C = \dfrac{2\varepsilon_0 A}{d\ln 2} = \dfrac{2C_0}{\ln 2} \]
Read it. invert, with \(C_0 = \varepsilon_0A/d\).
For the \(88.5\,\text{pF}\) capacitor, with \(\ln 2 = 0.693\), \(C\) \(= 2 \times 88.5/0.693\) \(\approx 255\,\text{pF}\). It lies between \(2C_0\) and \(4C_0\), as it must. More generally, \(K\) rising in a straight line from \(K_1\) to \(K_2\) across the gap gives \(C = C_0(K_2 - K_1)/\ln(K_2/K_1)\).
Now let \(K\) change along the plates instead, from \(2\) at one edge to \(4\) at the other. Cut the gap into thin strips across it, each from plate to plate. Every strip has the full voltage, so the strips are in parallel and their capacitances add. A strip of width \(dy\) has \(K(y)\varepsilon_0b\,dy/d\). Adding them gives \(C_0\) times the average of \(K\), \(3C_0 = 265.5\,\text{pF}\), the same as two halves side by side.
One question settles every varying case. Across the gap, the slices share one charge and are in series, so \(1/K\) is averaged. Along the plates, the strips share one voltage and are in parallel, so \(K\) itself is averaged. Averaging \(1/K\) always gives the smaller capacitance.
The dielectric constant rises in a straight line across the gap, from \(1\) at one plate to \(3\) at the other. The capacitor has \(50\,\text{pF}\) with air. What is its capacitance, in picofarads? Take \(\ln 3 = 1.10\).
Many fuel tanks, in aircraft and in industry, measure their level with a capacitor. Two coaxial metal tubes stand upright in the tank. The fuel between them is a dielectric, and air fills the rest. As the level rises the tubes become more filled, exactly like the slab sliding in. The capacitance rises in a straight line with the level, so a meter reading \(C\) can be marked in litres.
- Adding \(t\) and \(t/K\) the wrong way: \(C = \varepsilon_0A/(d - t/K)\).The slab replaces a thickness \(t\) of air by an effective \(t/K\): the gap acts as \(d - t + t/K\).
- Treating stacked layers as parallel because they fill the whole area.Stacked layers share one charge and are in series; side-by-side parts share one voltage and are in parallel.
- Averaging \(K\) across the gap.Across the gap the slices are in series, so \(1/K\) is averaged, through an integral.
- Expecting the slab to be pushed out when the battery stays connected.The battery supplies twice the rise in the store, so the slab is still pulled in.