## SAT Physics Subject Test

**Chapter 9 ****Electric Potential and Capacitance**

**CAPACITANCE**

You may see a question on capacitors on the SAT Physics Subject Test, so let’s discuss them now. Consider two conductors, separated by some distance, that carry equal but opposite charges, +*Q* and –*Q*. This pair of conductors make up a system called a **capacitor**. Work must be done to create this separation of charge, and, as a result, potential energy is stored. Capacitors are basically storage devices for electrical potential energy.

The most common conductors are parallel metal plates or sheets, and these capacitors are called **parallel-plate capacitors**. We’ll assume that the distance *d* between the plates is small compared to the dimensions of the plates, and in this case, the electric field between the plates is uniform.

The ratio of *Q* to the voltage, Δ *V*, for *any* capacitor, is called its **capacitance** (*C*).

*C* =

For a parallel-plate capacitor, we have

*C* =

Capacitance is a measure of the capacity for holding charge. The greater the capacitance, the more charge can be stored on the plates at a given potential difference. The capacitance of any capacitor depends only on the size, shape, and separation of the conductors. From the definition *C* =*Q*/*V*, the units of *C* are coulombs per volt. One coulomb per volt is renamed one **farad** (abbreviated F): 1 C/V = 1 F .

**Questions 6-7**

A 10-nanofarad parallel-plate capacitor holds a charge of magnitude 50 C on each plate.

6. What is the potential difference between the plates?

7. If the plates are separated by a distance of 0.885 mm, what is the area of each plate?

Here’s How to Crack It

6. From the definition, *C* = , we find that

7. From the equation *C* = , we can calculate the area, A, of each plate.