Irodov 3.200: The Source Takes Back Twice – Amit & Deepali Sinha Foundation

Problem Irodov 3.200

A metal plate, parallel to the plates of a parallel-plate capacitor, lies in the gap; its thickness is a fraction $\eta = 0.60$ of the gap width. Without the metal plate the capacitance would be $C = 20$ nF. The capacitor is connected to a source of constant voltage $V = 100$ V, and the metal plate is slowly pulled out of the gap. Find

the change in the energy of the capacitor;
the mechanical work done in pulling the plate out.

The capacitor with the metal plate, of thickness η d, being drawn out with velocity v, and the source V across the plates. — The capacitor with the metal plate, of thickness $\eta d$ , being drawn out with velocity $\vect{v}$ , and the source $V$ across the plates.

Given

$\eta = 0.60$
$C = 20$ nF (no metal plate)
$V = 100$ V
slow extraction
source connected throughout
metal plate covers the plates (assumed)

Find

(a) the energy change

\Delta W

· (b) the mechanical work

A

Solution

Step 1The capacitance with the metal plate in

Call the gap $d$ and the plate area $S$ , so that $C = \varepsilon_0S/d$ , and take the metal plate to cover the whole plate area. There is no field inside the metal, so the field fills only the remaining width $(1 - \eta)d$ ; the capacitor behaves as one of that gap:

C_1 = \frac{\varepsilon_0 S}{(1 - \eta)d} = \frac{C}{1 - \eta} \tag{1}

Pulling the plate out lowers the capacitance from $C_1$ to $C$ while the source holds the voltage at $V$ .

Step 2(a) The change in the capacitor’s energy

At fixed voltage the energy is $CV^2/2$ , so the change from $C_1$ to $C$ is

\Delta W = \frac{CV^2}{2} - \frac{C_1V^2}{2} = \frac{(C - C_1)V^2}{2} \tag{2}

Using (1), the difference of capacitances is

C - C_1 = C\left(1 - \frac{1}{1 - \eta}\right) = -\frac{C\eta}{1 - \eta} \tag{3}

Numerically:

\begin{aligned} CV^2 &= 20\times10^{-9}\cdot 100^2\ \text{J} = 2.0\times10^{-4}\ \text{J}\\[2pt] \Delta W &= -\frac{2.0\times10^{-4}}{2}\cdot\frac{0.60}{0.40}\ \text{J} \end{aligned}

\Delta W = -\frac{CV^2}{2}\,\frac{\eta}{1 - \eta} = -0.15\text{ mJ}

(4)

The capacitor loses energy: at fixed voltage, a smaller capacitance holds less.

Step 3The work done by the source

The charge on the capacitor falls from $C_1V$ to $CV$ , and that charge is driven back into the source against its voltage. The work done by the source is therefore negative:

A_{\mathrm{s}} = V\,\Delta q = (C - C_1)V^2 \tag{5}

Step 4(b) The mechanical work

The plate moves slowly, so it gains no kinetic energy and the current is too small to produce Joule heat. The energy given to the capacitor equals the work of the external agent plus the work of the source:

\Delta W = A + A_{\mathrm{s}}

With (2) and (5), then (3):

A = \Delta W - A_{\mathrm{s}} = \frac{(C - C_1)V^2}{2} - (C - C_1)V^2 = \frac{(C_1 - C)V^2}{2}

A = \frac{CV^2}{2}\,\frac{\eta}{1 - \eta} = 0.15\text{ mJ}

(6)

The work is positive: the field pulls the metal plate into the gap, so it has to be pulled out.

Check

Energy accounting with the numbers: the agent supplies $0.15$ mJ and the capacitor gives up $0.15$ mJ; the source takes in their sum, $0.30$ mJ. That is the voltage times the charge returned to it, with $C_1 = 50$ nF from (1):

(C_1 - C)V\cdot V = 30\times10^{-9}\cdot 100\cdot 100\ \text{J} = 0.30\ \text{mJ}

Limit: with $\eta \to 0$ the plate is infinitely thin and both answers vanish, as they must.

Remark

The mechanical work and the change of field energy are equal and opposite, the general rule for a capacitor held at a constant voltage: for any displacement $x$ of a body in it, the electric force along $x$ is $+\partial W/\partial x$ at fixed $V$ , not $-\partial W/\partial x$ , because the source’s work is always twice the change of the field energy. If the source were disconnected first, the charge would be fixed, the energy would rise by the factor $C_1/C = 2.5$ , and the work would show up entirely as that rise.