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# Electric Field of a Parallel Plate Capacitor

Difficulty Level: At Grade Created by: CK-12

Suppose we have two parallel metal plates set a distance d\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 E=Vd\begin{align*} E = \frac{V}{d} \end{align*} where V is the potential difference (voltage) between the two plates.

The amount of charge, Q\begin{align*}Q\end{align*}, held by each plate is given by Q=CV\begin{align*} Q = CV \end{align*} where again V\begin{align*}V\end{align*} is the voltage difference between the plates and C\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 C=ϵ0Ad\begin{align*} C = \frac{\epsilon_0 A}{d} \end{align*} where A\begin{align*}A\end{align*} is the area of the plates, d\begin{align*} d \end{align*} is the distance between the plates, and ϵ0\begin{align*} \epsilon_0 \end{align*} is the permittivity of free space whose value is 8.84×1012C/Vm\begin{align*} 8.84 \times 10^{-12} C/V \cdot m \end{align*}.

The electric field between the capacitor plates stores energy. The electric potential energy, UC\begin{align*}U_{\text{C}} \end{align*}, stored in the capacitor is given by

UC=12CV2\begin{align*} U_{\text{C}} = \frac{1}{2}CV^2 \end{align*}

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 chapter.

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.

## More on Electric and Gravitational Potential

There are several differences between our approach to gravity and electricity that could cause confusion. First, with gravity we usually used the concept of "energy", rather than "energy difference". Second, we spoke about it in absolute terms, rather than "per unit mass".

To address the first issue: when we dealt with gravitational potential energy we had to set some reference height h=0\begin{align*} h = 0 \end{align*} where it is equal to mg×0=0\begin{align*} mg \times 0 = 0 \end{align*}. In this sense, we were really talking about potential energy differences rather than absolute levels then also: at any point, we compared the gravitational potential energy of an object to the energy it would have had at the reference level h=0\begin{align*} h = 0 \end{align*}. When we used the formula Ug=mgΔh\begin{align*} U_g = mg\Delta h \end{align*} we implicitly set the initial point as the zero: no free lunch! For the same reason, we use the concept of electric potential difference between two points --- or we need to set the potential at some point to 0, and use it as a reference. This is not as easy in this case though; usually a point very far away ("infinitely" far) is considered to have 0 electric potential.

Regarding the second issue: in the chapter on potential energy, we could have gravitational potential difference between two points at different heights as gΔh\begin{align*} g\Delta h \end{align*}. This, of course, is the work required to move an object of mass one a height Δh\begin{align*} \Delta h \end{align*} against gravity. To find the work required for any other mass, we would multiply this by its magnitude. In other words, WWork=mMass×gΔhPotential Difference\begin{align*} \underbrace{W}_{\text{Work}} =\underbrace{m}_{\text{Mass}} \times \underbrace{g\Delta h}_{\text{Potential Difference}} \end{align*} Which is exactly analogous to the equation above.

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Date Created:
Feb 23, 2012