A capacitor is a device that stores charge. It is typically two very large flat plates wrapped in a cylinder with a dielectric substance in between the two. The dielectric allows the plates to be at a high voltage (and thus store more charge) without arcing between them. The capacitance, literally tells you how much charge it can hold. The capacitance of a capacitor only depends on its geometry. Here some important things to keep in mind in regards to capacitors:

- \begin{align*}Q\end{align*} refers to the amount of positive charge stored on the high voltage side of the capacitor; an equal and opposite amount, \begin{align*}-Q\end{align*}, of negative charge is stored on the low voltage side of the capacitor.

- Current can flow
*into*a capacitor from either side, but current doesn’t flow across the capacitor from one plate to another. The plates do not touch, and the substance in between is insulating, not conducting.

- One side of the capacitor fills up with negative charge, and the other fills up with positive charge. The reason for the thin, close plates is so that you can use the negative charge on one plate to attract and hold the positive charge on the other plate. The reason for the plates with large areas is so that you can spread out the charge on one plate so that its self-repulsion doesn’t stop you from filling it with more charge.

- Typical dielectric constants κ are roughly 5.6 for glass and 80 for water. What these “dielectric” substances do is align their electric polarity with the electric field in a capacitor (much like atoms in a magnetic material) and, in doing so, reduces the electric field for a given amount of charge. Thereby allowing for more charge to be stored for a given Voltage.

- The electrical circuit symbol for a capacitor is two flat plates, mimicking the geometry of a capacitor, which typically consists of two flat plates separated by a small distance. The plates are normally wrapped around several times to form a cylindrical shape.

#### Example

You create a simple capacitor by placing two .25m square metal plates .01m apart and then connecting each plate to one end of a battery. (a) If the battery can create a voltage drop of 12V, how much charge can be stored in the capacitor. (b) If you immersed the whole system in water, how much more charge could you store on the capacitor?

(a): To solve this problem, we can use the equations give above. First we'll find the capacitance of the two plates based on their dimensions. Since there is not dielectric between the plates to start, the dielectric constant is 1.

\begin{align*} C&=\frac{\kappa \epsilon_o A}{d}\\ C&=\frac{1*8.85*10^{-12}\:\text{F/m} * (.25\:\text{m})^2}{.01\:\text{m}}\\ C&=5.5*10^{-11}\:\text{F}\\ \end{align*}

Now we can find out how much charge can be stored on the capacitor.

\begin{align*} Q&=CV\\ Q&=5.5*10^{-11}\:\text{F} * 12\:\text{V}\\ Q&=6.63*10^{-10}\:\text{C}\\ \end{align*}

(b): The new capacitance will just be 80x the original capacitance because that is the dielectric constant of water. We can use this to calculate how much charge can be stored on the submerged capacitor.

\begin{align*} Q&=80CV\\ Q&=80*5.5*10^{-11}\:\text{F} * 12\:\text{V}\\ Q&=5.31*10^{-8}\:\text{F}\\ \end{align*}

As you can see, the capacitance of these plates is pretty small. Most capacitors have very thin, but very long metal sheets that are rolled up into a cylinder and separated by only a few millimeters; this provides a much greater capacitance in a much smaller volume.

### Simulation

Capacitor Lab (PhET Simulation)

### Review

- Design a parallel plate capacitor with a capacitance of \begin{align*}100\;\mathrm{mF}\end{align*}. You can select any area, plate separation, and dielectric substance that you wish.
- Show, by means of a sketch illustrating the charge distribution, that two identical parallel-plate capacitors wired in parallel act exactly the same as a single capacitor with twice the area.
- A certain capacitor can store \begin{align*}5\;\mathrm{C}\end{align*} of charge if you apply a voltage of \begin{align*}10\;\mathrm{V}\end{align*}.
- How many volts would you have to apply to store \begin{align*}50\;\mathrm{C}\end{align*} of charge in the same capacitor?
- Why is it harder to store more charge?

- A capacitor is charged and then unhooked from the battery. A dielectric is then inserted using an insulating glove.
- Does the electric field between plates increase or decrease? Why?
- did it take negative work, no work or positive work when the dielectric was inserted? (.e. did it get ‘sucked in’ or did you have to ‘push it in’ or neither) Explain.

The capacitor is now reattached to the battery

- Does the voltage of the capacitor increase, decrease or stay the same? Explain
- The dielectric is now removed from the middle with battery attached. What happens?

### Review (Answers)

- a. \begin{align*}100 \;\mathrm{V}\end{align*} b. Because as charges build up they repel each other from the plate and a greater voltage is needed to create a stronger electric field forcing charge to flow
- a. decrease b. negative work ('sucked in') c. Goes back up to what it was before inserted dielectric d. V is constant, since capacitance goes down, charge must go down. There's probably a discharge across the capacitor.