Irodov 3.141: Fixed Charge Versus Fixed Voltage – Amit & Deepali Sinha Foundation

Problem Irodov 3.141

A parallel-plate capacitor with air between its plates has plates of area $S$ . The plates are slowly pulled apart, their separation growing from $x_1$ to $x_2$ . How much work must be done if, during the process,

the charge on the capacitor is kept constant and equal to $q$ ;
the voltage across the capacitor is kept constant and equal to $V$ ?

The air capacitor edge on: two plates of area S a distance x apart, with x growing from x_1 to x_2. — The air capacitor edge on: two plates of area $S$ a distance $x$ apart, with $x$ growing from $x_1$ to $x_2$ .

Given

plate area $S$
separation $x_1 \to x_2$
(a) charge $q$ constant
(b) voltage $V$ constant

Find

the work

A

in each case

Solution

Step 1Capacitance and energy at separation

x

Neglecting edge effects, the capacitance of the air capacitor at separation $x$ is

C(x) = \frac{\varepsilon_0 S}{x} \tag{1}

The plates are moved slowly, so they gain no kinetic energy and the work done on them goes into the energy of the field and, in case (b), of the source.

Step 2(a) Constant charge

With the capacitor isolated, the only energy that changes is the field energy, $W = q^2/2C$ . Using (1):

W(x) = \frac{q^2}{2C} = \frac{q^2 x}{2\varepsilon_0 S}

The work equals the gain in this energy:

A_{\mathrm{a}} = W(x_2) - W(x_1) = \dfrac{q^2\,(x_2 - x_1)}{2\varepsilon_0 S}

(2)

Step 3(b) Constant voltage: energy of the field

Now the field energy is $W = CV^2/2$ , and it falls as the plates separate:

\Delta W = \frac{V^2}{2}\left[C(x_2) - C(x_1)\right] = \frac{\varepsilon_0 S V^2}{2}\left(\frac{1}{x_2} - \frac{1}{x_1}\right) \tag{3}

Step 4(b) Constant voltage: work of the source

The charge $CV$ also falls, so charge $\Delta q = V\Delta C$ (negative) passes through the source, which does work $V\Delta q$ on the system:

A_{\mathrm{src}} = V\,\Delta q = V^2\left[C(x_2) - C(x_1)\right] = \varepsilon_0 S V^2\left(\frac{1}{x_2} - \frac{1}{x_1}\right) \tag{4}

Step 5(b) The work done on the plates

Energy conservation: the work done by the external agent plus the work of the source equals the change in field energy. Using (3) and (4):

A_{\mathrm{b}} = \Delta W - A_{\mathrm{src}} = \frac{\varepsilon_0 S V^2}{2}\left(\frac{1}{x_2} - \frac{1}{x_1}\right) - \varepsilon_0 S V^2\left(\frac{1}{x_2} - \frac{1}{x_1}\right)

A_{\mathrm{b}} = \dfrac{\varepsilon_0 S V^2}{2}\left(\dfrac{1}{x_1} - \dfrac{1}{x_2}\right) = \dfrac{\varepsilon_0 S V^2\,(x_2 - x_1)}{2 x_1 x_2}

(5)

Check

Force check. Each plate sits in the field $E/2$ of the other, so it is pulled back with force $F = qE/2$ , and the agent must supply that force over the distance. In (a) the field $E = q/\varepsilon_0 S$ does not depend on $x$ ; in (b) the charge is $\varepsilon_0 SV/x$ and the field is $V/x$ :

\begin{aligned} A_{\mathrm{a}} &= \frac{q^2}{2\varepsilon_0 S}\,(x_2 - x_1)\\ A_{\mathrm{b}} &= \int_{x_1}^{x_2}\frac{\varepsilon_0 S V^2}{2x^2}\,dx = \frac{\varepsilon_0 S V^2}{2}\left(\frac{1}{x_1} - \frac{1}{x_2}\right) \end{aligned}

These are (2) and (5). Both works are positive, as they must be for plates that attract.

Remark

In (b) the agent does positive work and yet the field energy goes down: the agent’s work and the lost field energy both flow into the source, in the ratio one to one. This bookkeeping, $A_{\mathrm{src}} = 2\Delta W$ at constant voltage, holds for any slow change of geometry, and it is why the force on a conductor at fixed $V$ is $+\partial W/\partial x$ , the opposite sign of the fixed-charge rule $-\partial W/\partial x$ .