Irodov 3.201: Pulling Glass Against a Fixed Voltage – Amit & Deepali Sinha Foundation

Problem Irodov 3.201

A glass plate (permittivity $\varepsilon = 6.0$ ) completely fills the gap between the electrodes of a parallel-plate capacitor, whose capacitance without the glass would be $C = 20$ nF. The capacitor is connected to a source of constant voltage $V = 100$ V, and the glass plate is pulled out of the gap slowly and without friction. Find the change in the energy of the capacitor and the mechanical work done in pulling the plate out.

The capacitor with the glass plate of permittivity ε, which filled the gap, being drawn out with velocity v, and the source V across the plates. — The capacitor with the glass plate of permittivity $\varepsilon$ , which filled the gap, being drawn out with velocity $\vect{v}$ , and the source $V$ across the plates.

Given

$C = 20$ nF (no glass)
$V = 100$ V
glass, $\varepsilon = 6.0$ (table value)
slow, frictionless extraction
source connected throughout

Find

the energy change

\Delta W

· the mechanical work

A

Solution

Step 1The capacitance with the glass in

Glass has the permittivity $\varepsilon = 6.0$ (a table value). With the gap completely filled, every field strength between the plates is $\varepsilon$ times smaller at the same charge, so the capacitance is $\varepsilon$ times larger:

C_1 = \varepsilon C \tag{1}

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

Step 2The 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}

By (1), the difference of capacitances is

C - C_1 = C - \varepsilon C = -(\varepsilon - 1)C \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{(6.0 - 1)\cdot 2.0\times10^{-4}}{2}\ \text{J} \end{aligned}

\Delta W = -\frac{(\varepsilon - 1)CV^2}{2} = -0.50\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 4The mechanical work

The plate moves slowly and without friction, so it gains no kinetic energy, no heat is made at the surfaces, 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{(\varepsilon - 1)CV^2}{2} = 0.50\text{ mJ}

(6)

The work is positive: the field draws the glass into the gap, so it has to be pulled out.

Check

Energy accounting with the numbers: the agent supplies $0.50$ mJ and the capacitor gives up $0.50$ mJ; the source takes in their sum, $1.0$ mJ. That is the voltage times the charge returned to it, with $C_1 = 120$ nF from (1):

(C_1 - C)V\cdot V = 100\times10^{-9}\cdot 100\cdot 100\ \text{J} = 1.0\ \text{mJ}

Limit: with $\varepsilon = 1$ the plate is no different from air and both answers vanish; with $\varepsilon \to \infty$ the plate acts like a metal filling the whole gap, and both grow without bound, as $C_1$ does.

Remark

The force is felt only while the plate’s edge is inside the gap, where the fringing field at the end pulls the glass in; the totals above do not depend on how that force varies along the way. Had the source been disconnected before the extraction, the charge would stay fixed and the energy would rise by the factor $\varepsilon$ , from $\varepsilon CV^2/2$ to $\varepsilon^2CV^2/2$ ; the work, $(\varepsilon - 1)\varepsilon CV^2/2 = 3.0$ mJ, would then be $\varepsilon$ times larger.