# Chapter 16: Electric Potential

Difficulty Level: At Grade Created by: CK-12

If there is a strong enough electric field, sparks form in the air where the electricity jumps from one object to another.  In this chapter, we discuss electrical energy and electric potential.

Chapter Outline

### Chapter Summary

1. The electric field between two parallel-plate conductors is considered uniform far away from the plate edges if the size of the plates is large compared to their separation distance.
2. The potential energy of a charge \begin{align*}q\end{align*} at a point between two parallel-plate conductors is \begin{align*}PE=qEx\end{align*}, a reference point must be given such as \begin{align*}PE=0\end{align*} at \begin{align*}x=0\end{align*}.
3. A point charge \begin{align*}q\end{align*} has electric potential energy \begin{align*}PE_x\end{align*} and electric potential \begin{align*}V_x\end{align*} at point \begin{align*}x\end{align*}. Thus, \begin{align*}PE_x=qV_x\end{align*}
4. The word voltage is used when we mean potential difference.
5. It is common to write \begin{align*}V=Ed\end{align*}, where \begin{align*}V\end{align*} is understood to mean the voltage (or potential difference) between the plates of a parallel-plate conductor and \begin{align*}d\end{align*} is the distance between the plates.
6. The work done by the electric field in moving a charge between two parallel plate conductors is \begin{align*}W_{field}=-q \Delta V\end{align*}. The work done by an external force is \begin{align*}W_{external \ force}=q \Delta V\end{align*}.
7. Voltage can be thought of as the work per unit charge \begin{align*}V=\frac{W}{q}\end{align*}; that is, how much work is required per unit charge to move a charged particle in an electric field.
8. Capacitance of an air-gap capacitor is given by \begin{align*}C = \varepsilon_0 \frac{A}{d}\end{align*} where \begin{align*}A\end{align*} is the area of the capacitor and \begin{align*}d\end{align*} is the separation distance between the plates.
9. The charge on a capacitor is directly proportional to the voltage of the capacitor \begin{align*}Q = CV\end{align*}.
10. A dielectric material placed between the plates increases the capacitance of the capacitor. The capacitance of a capacitor with a dielectric is expressed as \begin{align*}C = k\varepsilon_0 \frac{A}{d}\end{align*}, where \begin{align*}k\end{align*} is the dielectric constant.
11. The energy stored in a capacitor can be expressed as

\begin{align*}U & = \frac{1}{2}QV\\ U & = \frac{1}{2}QV^2\\ U & = \frac{1}{2} \frac{Q^2}{C}\end{align*}
Where \begin{align*}Q\end{align*} is the charge on the capacitor and \begin{align*}V\end{align*} is the voltage of the capacitor.

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Jun 27, 2013