Irodov 3.142: Work to Pull Out a Slab – Amit & Deepali Sinha Foundation

Problem Irodov 3.142

A parallel-plate capacitor has, between its plates and parallel to them, a slab whose thickness is a fraction $\eta = 0.60$ of the gap width. Without the slab the capacitance would be $c = 20$ nF. With the slab in place, the capacitor is connected to a source of constant voltage $V = 200$ V, then disconnected, and then the slab is slowly pulled out of the gap. How much work is done in removing the slab if it is

made of metal;
made of glass (permittivity $\varepsilon = 6.0$ )?

(a) A metal slab of thickness η d in the gap d. — (a) A metal slab of thickness $\eta d$ in the gap $d$ .

(b) The same with a glass slab of permittivity ε. — (b) The same with a glass slab of permittivity $\varepsilon$ .

Given

$\eta = 0.60$
$c = 20$ nF (no slab)
$V = 200$ V, then disconnected
glass, $\varepsilon = 6.0$ (table value)
slab covers the plates (assumed)

Find

the work

A

for (a) a metal and (b) a glass slab

Solution

Step 1The charge is fixed at disconnection

Let $C_1$ be the capacitance with the slab in place. While connected, the capacitor takes the charge $C_1V$ ; once disconnected it keeps it while the slab is removed:

q = C_1 V \tag{1}

Step 2The capacitance with the slab in the gap

(a) With the metal slab: no field inside it; the lines end on its induced charge.

(b) With the glass slab: the field inside is ε times weaker. — (b) With the glass slab: the field inside is $\varepsilon$ times weaker.

Call the gap $d$ and the plate area $S$ , so that $c = \varepsilon_0S/d$ , and take the slab to cover the whole plate area. The slab and the two air layers, of total thickness $(1 - \eta)d$ , are in series: the voltage across the gap is the sum of $E\,(1-\eta)d$ in the air and $(E/\varepsilon)\,\eta d$ in the slab, where $E = q/\varepsilon_0 S$ is the field in the air.

V = \frac{q}{\varepsilon_0 S}\left[(1 - \eta)d + \frac{\eta d}{\varepsilon}\right]

Dividing $q$ by this voltage and writing $\varepsilon_0 S/d$ as $c$ :

C_1 = \frac{q}{V} = \frac{c}{1 - \eta + \eta/\varepsilon} = \frac{c\,\varepsilon}{\varepsilon - \eta(\varepsilon - 1)} \tag{2}

A metal slab has no field inside, which is the limit $\varepsilon \to \infty$ :

C_1^{\mathrm{metal}} = \frac{c}{1 - \eta} \tag{3}

Step 3Work as a gain in field energy

The capacitor is isolated and the slab moves slowly, so the work done equals the rise in field energy from $q^2/2C_1$ to $q^2/2c$ . With (1):

A = \frac{q^2}{2c} - \frac{q^2}{2C_1} = \frac{C_1^2V^2}{2}\left(\frac{1}{c} - \frac{1}{C_1}\right)

A = \frac{C_1V^2\,(C_1 - c)}{2c}

(4)

The work is positive because $C_1 > c$ : the field pulls the slab into the gap, and pulling it out takes work.

Step 4(a) Metal slab

Substituting (3) into (4):

\begin{aligned} C_1 - c &= \frac{c}{1 - \eta} - c = \frac{c\,\eta}{1 - \eta}\\ A_{\mathrm{a}} &= \frac{1}{2c}\cdot\frac{cV^2}{1 - \eta}\cdot\frac{c\,\eta}{1 - \eta} \end{aligned}

Numerically:

\begin{aligned} cV^2 &= 20\times10^{-9}\cdot 200^2\ \text{J} = 8.0\times10^{-4}\ \text{J}\\ A_{\mathrm{a}} &= \frac{8.0\times10^{-4}\cdot 0.60}{2\cdot 0.40^2}\ \text{J} \end{aligned}

A_{\mathrm{a}} = \dfrac{cV^2\,\eta}{2(1 - \eta)^2} \approx 1.5\text{ mJ}

(5)

Step 5(b) Glass slab

Glass has $\varepsilon = 6.0$ (a table value). Substituting (2) into (4):

\begin{aligned} C_1 - c &= \frac{c\,\varepsilon}{\varepsilon - \eta(\varepsilon - 1)} - c = \frac{c\,\eta(\varepsilon - 1)}{\varepsilon - \eta(\varepsilon - 1)}\\ A_{\mathrm{b}} &= \frac{1}{2c}\cdot\frac{c\,\varepsilon V^2}{\varepsilon - \eta(\varepsilon - 1)}\cdot\frac{c\,\eta(\varepsilon - 1)}{\varepsilon - \eta(\varepsilon - 1)} \end{aligned}

Numerically, the bracket is $6.0 - 0.60\cdot 5.0 = 3.0$ , and

A_{\mathrm{b}} = \frac{8.0\times10^{-4}\cdot 0.60\cdot 6.0\cdot 5.0}{2\cdot 3.0^2}\ \text{J}

A_{\mathrm{b}} = \dfrac{cV^2\,\eta\,\varepsilon(\varepsilon - 1)}{2\left[\varepsilon - \eta(\varepsilon - 1)\right]^2} \approx 0.80\text{ mJ}

(6)

Check

Letting $\varepsilon \to \infty$ in (6), the bracket tends to $\varepsilon(1 - \eta)$ and the factor $\varepsilon(\varepsilon - 1)$ to $\varepsilon^2$ , so $A_{\mathrm{b}} \to cV^2\eta/2(1 - \eta)^2$ , the metal result (5). With $\varepsilon = 1$ the slab is air and $A_{\mathrm{b}} = 0$ , as it must. The order $A_{\mathrm{a}} > A_{\mathrm{b}}$ is right too: the metal shrinks the effective gap more, so it is held more strongly.

Remark

The work done shows up as field energy: the energy grows by the factor $C_1/c$ , here $2.5$ for the metal and $2.0$ for the glass, while the charge stays put and the voltage rises by the same factor. Had the source stayed connected, the voltage would be fixed, the energy would fall as the slab left, and the source would absorb both that drop and the work done, the bookkeeping of constant-voltage problems.