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A parallel-plate capacitor has square plates of side a = 10.0 cm, held horizontal and 2.0 mm apart…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A parallel-plate capacitor has square plates of side $a = 10.0\ \text{cm}$ , held horizontal and $2.0\ \text{mm}$ apart in air. A glass slab of dielectric constant $5.0$ is as long and as wide as the plates, and exactly fills the gap between them in thickness. Its mass is $40\ \text{g}$ , and it can slide along the gap without friction, as in the figure. Ignore the spreading of the field at the edges of the plates when you work out a capacitance. Take $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$ .

Side view of a parallel-plate capacitor: two horizontal plates, 10.0 cm long and 2.0 mm apart (the gap is drawn much larger than it is). A glass slab of dielectric constant 5.0, as long as the plates and exactly as thick as the gap, enters the gap from the left. A length x0 = 4.0 cm of it is inside the plates and the rest sticks out to the left. Wires join the two plates to a 300 V battery.

The slab is held with a length of $x_0 = 4.0\ \text{cm}$ inside the plates, and the capacitor is joined to a $300\ \text{V}$ battery.

What is the capacitance now?
The slab is released. What is the size of the force that pulls it in? Show that, while the battery stays joined, this force does not depend on how far the slab is inside.
How fast is the slab moving at the moment it is fully inside?
How much energy does the battery give out during that motion?

The experiment is repeated from the same start, but the battery is taken away just before the slab is released, so the charge on the plates stays fixed.

How fast is the slab moving now at the moment it is fully inside?

A student says the slab cannot speed up while the battery is joined, because the energy stored in the capacitor rises as the slab goes in. Your answers should show where the student goes wrong.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Write the capacitance as a function of the length $x$ inside. Then use energy: with the battery joined, the battery's work splits equally between the field and the slab; with the charge fixed, the field's energy alone pays for the slab's motion.

Step 1. Capacitance (a). $C(x) = \dfrac{\varepsilon_0 a}{d}\left[Kx + (a - x)\right] = \dfrac{\varepsilon_0 a}{d}\left[a + (K - 1)x\right]$ . At $x_0$ : $C = 115.1\ \text{pF}$ . (Empty it would be $44.25\ \text{pF}$ ; full, $221.25\ \text{pF}$ .)

Step 2. Force (b). At fixed $V$ the force is $F = \dfrac{V^2}{2}\dfrac{dC}{dx} = \dfrac{\varepsilon_0 a(K - 1)V^2}{2d}$ , with no $x$ in it:

\displaystyle F = \frac{8.85 \times 10^{-12}\ \text{F/m} \times 0.10\ \text{m} \times 4.0 \times (300\ \text{V})^2}{2 \times 2.0 \times 10^{-3}\ \text{m}} = 79.7\ \mu\text{N}

Step 3. Speed, battery joined (c). The force is constant over the remaining $6.0\ \text{cm}$ , so the kinetic energy is $79.7\ \mu\text{N} \times 0.06\ \text{m} = 4.78\ \mu\text{J}$ and $v = \sqrt{2KE/m} = 1.55\ \text{cm/s}$ .

Step 4. The battery (d). $W_{\text{bat}} = V^2\Delta C = 9.56\ \mu\text{J}$ . The stored energy rises by $\tfrac12V^2\Delta C = 4.78\ \mu\text{J}$ , and the rest, $4.78\ \mu\text{J}$ , is the slab's kinetic energy. This answers the student: the battery pays for both.

Step 5. Charge fixed (e). $Q = C(x_0)V = 34.5\ \text{nC}$ . The slab gains the fall in stored energy:

\displaystyle KE = \frac{Q^2}{2}\left(\frac{1}{C(x_0)} - \frac{1}{C(a)}\right) = 2.49\ \mu\text{J}

so $v = 1.11\ \text{cm/s}$ . The pull now weakens as the slab goes in, because the voltage falls.

Answer: (a) $115.1\ \text{pF}$ . (b) $79.7\ \mu\text{N}$ . (c) $1.55\ \text{cm/s}$ . (d) $9.56\ \mu\text{J}$ . (e) $1.11\ \text{cm/s}$ .

Check. At the start both cases have the same charge and voltage, so the same force. With the charge fixed, the force at full insertion has fallen to $21.5\ \mu\text{N}$ , so the slab gains less energy and moves more slowly.

Watch out. Writing the force as $-dU/dx$ with $U = \tfrac12CV^2$ gives the right size but the wrong direction (it would push the slab out). At fixed $V$ the battery's work must be counted; the force is $+\tfrac12V^2\,dC/dx$ .

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