Physics · Ch 2 — Electrostatic Potential and Capacitance

The Parallel Plate Capacitor With and Without a Dielectric

2.13

The Parallel Plate Capacitor With and Without a Dielectric

The parallel plate capacitor -- two identical, flat conducting plates, each of area $A$ , held parallel to each other a small distance $d$ apart -- is both the simplest capacitor geometry to analyse exactly and, in slightly disguised forms, the geometry underlying most practical capacitors.

Capacitance without a dielectric (vacuum or air gap). Let the plates carry charge $+Q$ and $-Q$ , so each plate carries a uniform surface charge density of magnitude $\sigma = Q/A$ . Provided the plate separation $d$ is small compared to the plates' own linear dimensions (so edge effects can be neglected), the field between two oppositely, uniformly charged plane sheets is uniform and has magnitude

E_0 = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}

directed straight from the positive plate to the negative plate. Using $V_0 = E_0 d$ (Section 2.3, the uniform-field special case) for the potential difference between the plates,

V_0 = \frac{Qd}{\epsilon_0 A} \quad \Longrightarrow \quad C_0 = \frac{Q}{V_0} = \frac{\epsilon_0 A}{d}

This is the standard result for a parallel plate capacitor with vacuum (or, to an excellent approximation, air) between the plates: capacitance increases with larger plate area $A$ (more surface to hold charge for the same field) and decreases with larger separation $d$ (a weaker field, hence a larger $V$ , for the same charge).

Capacitance with a dielectric filling the entire gap. If a dielectric of dielectric constant $K$ is inserted to fill the ENTIRE gap between the plates, the net field between the plates is reduced by exactly the factor $K$ found in Section 2.11: $E = E_0/K = Q/(K\epsilon_0 A)$ . The new potential difference is $V = Ed = Qd/(K\epsilon_0 A)$ , giving

C = \frac{Q}{V} = \frac{K\epsilon_0 A}{d} = K\,C_0

so filling the gap completely with a dielectric of constant $K$ increases the capacitance by exactly that same factor $K$ -- the reduced field means a smaller voltage is needed for the same charge, and a smaller voltage for the same charge is precisely what a LARGER capacitance means.

A dielectric slab only partly filling the gap. If a dielectric slab of thickness $t < d$ (and the same constant $K$ ) is inserted so that it fills only part of the gap, leaving a remaining air gap of thickness $d - t$ , the total potential difference is the SUM of the potential drop across the air portion (field $E_0$ , thickness $d-t$ ) and across the dielectric portion (field $E_0/K$ , thickness $t$ ): …

Figure 1Parallel plate capacitor with a dielectric slab between the plates

What this figure shows. Two identical rectangular conducting plates, each of area $A$ , are drawn parallel to each other, separated by a distance $d$ , oriented vertically with a small gap of air visible between them. The left plate carries a uniform row of $+$ signs along its inner face and is connected by a wire to the positive terminal of a battery symbol drawn to the left; the right plate carries a uniform row of $-$ signs along its inner face and is connected by a wire to the battery's negative terminal, with the battery's own potential difference labelled $V$ . A shaded rectangular slab, drawn distinctly (a different fill pattern from the plates) and labelled with dielectric constant $K$ , is shown filling the entire gap between the two plates, touching both inner faces, with a thin double-headed arrow drawn across the gap labelled $d$ to mark the plate separation and a second double-headed arrow drawn along one plate's edge labelled with the plate area $A$ . A short inset to one side of the main figure repeats the same two plates and the same separation $d$ , but WITHOUT the shaded dielectric slab …

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