Irodov 3.145: The Pull That Ignores Depth – Amit & Deepali Sinha Foundation

Problem Irodov 3.145

A cylindrical capacitor has electrodes of mean radius $R$ separated by a gap $d$ , with $d \ll R$ , and is held at a constant voltage $V$ . A cylindrical sleeve of dielectric of permittivity $\varepsilon$ , whose thickness equals the gap so that it fills the space between the electrodes wherever it is present, is partly inserted between them. Find the magnitude of the electric force that pulls the sleeve into the capacitor.

The capacitor in axial section: electrodes of mean radius R, gap d R (drawn wider for clarity), and the dielectric sleeve ε partly inserted from the right. — The capacitor in axial section: electrodes of mean radius $R$ , gap $d \ll R$ (drawn wider for clarity), and the dielectric sleeve $\varepsilon$ partly inserted from the right.

Given

permittivity $\varepsilon$
mean radius $R$
gap $d \ll R$
constant voltage $V$

Find

the force

F

pulling the dielectric in

Solution

Step 1Capacitance with the sleeve in to a depth

x

Because $d \ll R$ the gap is a parallel-plate capacitor rolled into a tube: a strip of length $dl$ along the axis has plate area $2\pi R\,dl$ and capacitance $\varepsilon_0\varepsilon\,2\pi R\,dl/d$ . Let the electrodes have length $l$ and the sleeve be in to a depth $x$ ; the filled and empty parts are in parallel:

C(x) = \frac{2\pi\varepsilon_0 R}{d}\left[\varepsilon x + (l - x)\right] \tag{1}

Step 2The energy balance at constant voltage

Let the sleeve move in by $dx$ , slowly. The capacitance grows by $dC$ , the source pushes charge $V\,dC$ through itself and does work $V^2dC$ , and the field energy $CV^2/2$ grows by $V^2dC/2$ . The rest is the work $F\,dx$ done by the electric force on the sleeve:

\begin{aligned} V^2\,dC &= d\left(\frac{CV^2}{2}\right) + F\,dx = \frac{V^2}{2}\,dC + F\,dx\\ F &= \frac{V^2}{2}\,\frac{dC}{dx} \end{aligned}

Step 3The force

At constant voltage the field has the same strength in the filled and the empty part, while the charge on the electrodes is denser over the sleeve (ε times; the glyph rows there are drawn at their densest pitch). The electric force F pulls the sleeve in. — At constant voltage the field has the same strength in the filled and the empty part, while the charge on the electrodes is denser over the sleeve ( $\varepsilon$ times; the glyph rows there are drawn at their densest pitch). The electric force $\vect F$ pulls the sleeve in.

Differentiating (1) gives a constant, and substituting it:

\begin{aligned} \frac{dC}{dx} &= \frac{2\pi\varepsilon_0 R}{d}\,(\varepsilon - 1)\\ F &= \frac{V^2}{2}\cdot\frac{2\pi\varepsilon_0 R\,(\varepsilon - 1)}{d} \end{aligned}

F = \dfrac{\pi\varepsilon_0(\varepsilon - 1)R\,V^2}{d}

(2)

It is positive, so the force draws the sleeve into the gap.

Check

With $\varepsilon = 1$ the sleeve is just air and $F = 0$ . Unrolling the tube gives a flat capacitor of width $b = 2\pi R$ and gap $d$ , for which the same argument gives the familiar pull $\varepsilon_0(\varepsilon - 1)bV^2/2d$ on a slab; with $b = 2\pi R$ this is (2). Dimensions: $\varepsilon_0 V^2$ is F/m times V $^2$ , that is J/m, and $R/d$ is a pure number, so $F$ is in J/m = N.

Remark

The force does not depend on how far the sleeve is in: each extra length $dx$ adds the same capacitance. It also grows as $1/d$ at a fixed voltage, because the field $V/d$ and the energy density both grow as the gap narrows, which is why a dielectric liquid climbs strongly between closely spaced electrodes held at a voltage.