Suppose we have two parallel metal plates set a distance \begin{align*}d\end{align*} from one another. We place a positive charge on one of the plates and a negative charge on the other. In this configuration, there will be a uniform electric field between the plates pointing from, and normal to, the plate carrying the positive charge. The magnitude of this field is given by \begin{align*} E = \frac{V}{d} \end{align*} where V is the potential difference (voltage) between the two plates.

The amount of charge, \begin{align*}Q\end{align*}, held by each plate is given by \begin{align*} Q = CV \end{align*} where again \begin{align*}V\end{align*} is the voltage difference between the plates and \begin{align*}C\end{align*} is the capacitance of the plate configuration. Capacitance can be thought of as the capacity a device has for storing charge . In the parallel plate case the capacitance is given by \begin{align*} C = \frac{\epsilon_0 A}{d} \end{align*} where \begin{align*}A\end{align*} is the area of the plates, \begin{align*} d \end{align*} is the distance between the plates, and \begin{align*} \epsilon_0 \end{align*} is the permittivity of free space whose value is \begin{align*} 8.84 \times 10^{-12} C/V \cdot m \end{align*}.

The electric field between the capacitor plates stores energy.

Where does this energy come from? Recall, that in our preliminary discussion of electric forces we assert that "like charges repel one another". To build our initial configuration we had to place an excess of positive and negative charges, respectively, on each of the metal plates. Forcing these charges together on the plate had to overcome the mutual repulsion that the charges experience; this takes work. The energy used in moving the charges onto the plates gets stored in the field between the plates. It is in this way that the capacitor can be thought of as an energy storage device. This property will become more important when we study capacitors in the context of electric circuits in the next several Concepts.

Note: Many home-electronic circuits include capacitors; for this reason, it can be dangerous to mess around with old electronic components, as the capacitors may be charged even if the unit is unplugged. For example, old computer monitors (not flat screens) and TVs have capacitors that hold dangerous amounts of charge hours after the power is turned off.

\begin{align*} U_{\text{C}} = \frac{1}{2}CV^2 \end{align*}

#### Example

You have a capacitor with capacitance \begin{align*}100\text{nF}\end{align*}. If you connected it to a 12V battery, how much energy is stored in the capacitor when it is fully charged? If you were to submerge this capacitor in water, how much energy could be stored in it?

We'll just use the equation given above to calculate the energy stored on the capacitor.

\begin{align*} U_c&=\frac{1}{2}CV^2\\ U_c&=\frac{1}{2}100*10^{-9}\:\text{F}*12\:\text{V}\\ U_c&=6*10^{-7}\:\text{J}\\ \end{align*}

By adding a dielectric, we increase the capacitance of the capacitor by a factor of the dielectric constant. The dielectric constant of water is 80, so the new capacitance will be 80 times the original capacitance.

\begin{align*} U_c&=\frac{1}{2}80CV^2\\ U_c&=\frac{1}{2}80*100*10^{-9}\:\text{F}*12\:\text{V}\\ U_c&=4.8*10^{-5}\:\text{J}\\ \end{align*}

### Interactive Simulation

### Review

- You have a \begin{align*}5 \mu \mathrm{F}\end{align*} capacitor.
- How much voltage would you have to apply to charge the capacitor with \begin{align*}200\;\mathrm{C}\end{align*} of charge?
- Once you have finished, how much potential energy are you storing here?
- If all this energy could be harnessed to lift 100 lbs. into the air, how high would you be lifted?

- A certain capacitor can store \begin{align*}500\;\mathrm{J}\end{align*} of energy (by storing charge) if you apply a voltage of \begin{align*}15\;\mathrm{V}\end{align*}. How many volts would you have to apply to store \begin{align*}1000\;\mathrm{J}\end{align*} of energy in the same capacitor? (Important: why isn’t the answer to this just \begin{align*}30\;\mathrm{V}\end{align*}?)
- Marciel, a bicycling physicist, wishes to harvest some of the energy he puts into turning the pedals of his bike and store this energy in a capacitor. Then, when he stops at a stop light, the charge from this capacitor can flow out and run his bicycle headlight. He is able to generate \begin{align*}18\;\mathrm{V}\end{align*} of electric potential, on average, by pedaling (and using magnetic induction). If Mars wants to provide \begin{align*}0.5\end{align*} A of current for 60 seconds at a stop light, what should the capacitance of his capacitor be?

### Review (Answers)

- a. \begin{align*}4 \times 10^7 \;\mathrm{V}\end{align*} b. \begin{align*}4 \times 10^9 \;\mathrm{J}\end{align*} c. About \begin{align*} 9000 \;\mathrm{km}\end{align*}
- \begin{align*}21 \;\mathrm{V}, \;\mathrm{V}\end{align*} is squared.
- \begin{align*}3.3 \;\mathrm{F}\end{align*}