Capacitors.

Step 1 of 5

Two plates and a battery

A capacitor is two conductors with an insulator between them. Here they are two square aluminium plates, with air between them, connected to a 9 V battery. No current passes across the gap itself. The moment it is connected, the battery pulls electrons off one plate and pushes as many electrons onto the other, until the voltage between the plates equals its own. Then the current stops.

Show charge and field has cut the plates open and shows the charge on the sides that face each other. Answer the question from what you see.

Step 2 of 5

Capacitance

How much charge the plates take for each volt is set by the capacitor’s capacitance:

$C = \frac{Q}{V}$⁠

$Q$⁠ is the charge on each plate. Capacitance is measured in farads, 1 F = 1 C/V, and a farad is a great deal, so small capacitors are measured in picofarads, $1\ \mathrm{pF} = 10^{-12}\ \mathrm{F}$$1\ \mathrm{pF} =$$10^{-12}\ \mathrm{F}$⁠. For a parallel-plate capacitor, the capacitance depends only on its shape and its material:

$C = \kappa \varepsilon_0 \frac{A}{d}, \qquad \varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{F/m}$$C =$$\kappa \varepsilon_0 \frac{A}{d}, \qquad \varepsilon_0 =$$8.85 \times 10^{-12}\ \mathrm{F/m}$⁠

$A$⁠ is the area of each plate in square metres, $d$⁠ the gap in metres, and $\kappa$⁠ the dielectric constant of what lies between the plates: almost exactly $1$⁠ for air. Bigger plates hold more charge, and plates farther apart hold less.

The readings beside the Scene are hidden now. Work out the capacitance.

Step 3 of 5

Storing a charge

The readings are back. Change Plate area and Gap between the plates until each plate carries a charge of 100 pC, and press Check. The battery is still 9 V, so $Q = CV$⁠.

Step 4 of 5

Between the plates

Now we put an insulating material between the plates, with the battery still connected. The material does not conduct, but the charges in its molecules shift a little in the field: it is polarised, like the scraps of paper under the balloon. Predict what will happen to the charge on the plates, then watch: the Scene will slide in Glass and bring the readings back.

Step 5 of 5

The energy in a capacitor

To move charge from plate to plate the battery does work, and that energy is stored in the field between the plates. At first the plates are empty, and the voltage between them rises in a straight line with the charge, from $0$⁠ to $V$⁠. So the average voltage the charge is moved across is $\dfrac{V}{2}$⁠, and the energy stored is the area of the triangle under the graph of voltage against charge:

$U = \frac12 Q V = \frac12 C V^2$$U =$$\frac12 Q V =$$\frac12 C V^2$⁠

The capacitance of these plates, 400 cm² at 1.5 cm in air, is about 23.6 pF. How much energy do they store on a 9 V battery?