Irodov 3.146: A Constant Torque on a Turned Plate – Amit & Deepali Sinha Foundation

Problem Irodov 3.146

A capacitor is made of two fixed parallel plates, each a semicircle of radius $R$ , one above the other. Between them is a movable semicircular plate of radius $R$ made of dielectric of permittivity $\varepsilon$ , which can turn about the axis $O$ through the common center, perpendicular to the plates. Its thickness $d$ is practically equal to the separation of the fixed plates. A voltage $V$ is applied across the capacitor. The dielectric plate is turned about $O$ by an angle $\varphi$ from the position in which it exactly covers the fixed plates, so that a sector of angle $\varphi$ of it sticks out of the gap. Find the magnitude of the moment of the electric forces about $O$ acting on the dielectric plate.

Top view. The fixed semicircular plates (shaded) over the diameter, the axis O, and the dielectric plate turned by φ: its hidden edges are dashed, and the sector of angle φ that sticks out is shown below the diameter. — Top view. The fixed semicircular plates (shaded) over the diameter, the axis $O$ , and the dielectric plate turned by $\varphi$ : its hidden edges are dashed, and the sector of angle $\varphi$ that sticks out is shown below the diameter.

Given

plate radius $R$
dielectric $\varepsilon$ , thickness $d$ (equal to the gap)
voltage $V$
turned by $\varphi$

Find

the magnitude of the torque

N

about

O

Solution

Step 1Capacitance as a function of the angle

With the plate turned by $\varphi$ , the gap holds a dielectric sector of angle $\pi - \varphi$ and an empty sector of angle $\varphi$ , each a parallel-plate capacitor of gap $d$ , connected in parallel. A sector of angle $\alpha$ has area $R^2\alpha/2$ , so, neglecting edge effects,

C(\varphi) = \frac{\varepsilon_0}{d}\left[\varepsilon\,\frac{R^2(\pi - \varphi)}{2} + \frac{R^2\varphi}{2}\right] = \frac{\varepsilon_0R^2}{2d}\left[\varepsilon\pi - (\varepsilon - 1)\varphi\right] \tag{1}

Step 2Torque from the energy at constant voltage

Let the plate turn slowly by $d\varphi$ at the fixed voltage $V$ . The source does work $V^2dC$ , the field energy rises by $V^2dC/2$ , and the remainder is the work $N\,d\varphi$ of the electric torque on the plate:

\begin{aligned} V^2\,dC &= \frac{V^2}{2}\,dC + N\,d\varphi\\ N &= \frac{V^2}{2}\,\frac{dC}{d\varphi} \end{aligned}

Here $N$ counts positive in the sense of increasing $\varphi$ .

Step 3The torque

Differentiating (1) and substituting:

\begin{aligned} \frac{dC}{d\varphi} &= -\frac{\varepsilon_0R^2(\varepsilon - 1)}{2d}\\ N &= -\frac{V^2}{2}\cdot\frac{\varepsilon_0R^2(\varepsilon - 1)}{2d} \end{aligned}

|N| = \dfrac{(\varepsilon - 1)\,\varepsilon_0R^2V^2}{4d}

(2)

The torque N on the dielectric plate turns it back into the gap, against φ. — The torque $N$ on the dielectric plate turns it back into the gap, against $\varphi$ .

The minus sign means the torque acts to decrease $\varphi$ : it turns the dielectric back into the gap, where the capacitance is larger.

Check

With $\varepsilon = 1$ the movable plate is electrically air and $N = 0$ . A check by force: the edge of the dielectric is a strip of length $R$ in the gap, and an element $dr$ of it at radius $r$ is pulled sideways like a slab entering a flat capacitor, with force $\varepsilon_0(\varepsilon - 1)V^2dr/2d$ ; its moment arm is $r$ , and

N = \int_0^R \frac{\varepsilon_0(\varepsilon - 1)V^2}{2d}\,r\,dr = \frac{\varepsilon_0(\varepsilon - 1)V^2R^2}{4d},

which is (2).

Remark

The torque is the same at every angle between $0$ and $\pi$ , because each degree of turn moves the same sector area in or out of the gap. Released from any such position, the plate is therefore turned back with constant torque, the rotary twin of the constant force on a slab that is being drawn into a flat capacitor. Since $N$ grows as $V^2$ whatever the sign of $V$ , a spring balancing it would turn the plate into a voltmeter for alternating as well as steady voltage.