**Problem 1—**Let’s get back to basics and talk about the operation of a capacitor. Suppose you have two large, flat plates that are close to each other (with respect to their diameter). If you charge them up to a given voltage, and then physically move the plates away from each other, what happens to the voltage? What happens to the strength of the electric field between them?

**Answer 1—**The capacitance of the plates drops with increasing distance, so the voltage between them rises, because the charge doesn’t change and the voltage is equal to the charge divided by the capacitance. At first, while the plate spacing is still small relative to their diameter, The capacitance is proportional to the inverse of the spacing, so the voltage rises linearly with the spacing. However, as the spacing becomes larger, the capacitance drops more slowly and the voltage rises at a lower rate as well.

While the plate spacing is small, the electric field is almost entirely directly between the two plates, with only minor “fringing” effects at the edges. Since the voltage rise is proportional to the distance in this regime, the electric field (e.g., in volts per meter) remains essentially constant. However, once the plate spacing becomes comparable to the diameter of the plates, and fringing effects begin to dominate, the field begins to spread out and weaken. Ultimately, at very large distances, at which the plates themselves can be considered points, the voltage is essentially constant, and the field strength directly between them becomes proportional to the inverse of the distance.

**Problem 2—**If you double the spacing between the plates of a charged capapcitor, the capacitance is cut in half, and the voltage is doubled. However, the energy stored in the capacitor is defined to be E = 0.5 C V^{2}. This means that at the wider spacing, the capacitor has twice the energy that it had to start with. Where did the extra energy come from?

**Answer 2—**There is an attractive force between the plates of a capacitor created by the electric field. Physically moving the plates apart requires doing work against this force, and this work becomes the additional potential energy that is stored in the capacitor.

**Question 3—**What happens when a dielectric is placed in an electric field? Why does the capacitance of pair of plates increase when the space betwenn them is filled with a dielectric?

**Answer 3—**Dielectric materials are made of atoms, and the atoms contain both positive and negative charges. Although neither the positive nor the negative charges are free to move about in the material (which is what makes it an insulator), they can be shifted to varying degress with respect to each other. An electric field causes this shift, and the shift in turn creates an opposing field that partially cancels the original field. Part of the field’s energy is absorbed by the dielectric.

In a capacitor, the energy absorbed by the dielectric reduces the field between the plates, and therefore reduces the voltage that is created by a given amount of charge. Since capacitance is defined to be the charge divided by the voltage, this means that the capacitance is higher with the dielectric than without it.

**Problem 4—**What is the piezoelectric effect?

**Answer 4—**With certain dielectrics, most notably quartz and certain ceramics, the displacement of charge also causes a significant mechanical strain (physical movement) of the crystal lattice. This effect works two ways — a physical strain also causes a shift in electric charges, creating an electric field. This effect can be exploited in a number of ways, including transducers for vibration and sound (microphones and speakers), as well as devices that have a strong mechanical resonance (e.g., crystals) that can be used to create oscillators and filters.

Contributed by David Tweed