Like gravity, the electric force can do work and has a potential energy associated with it. But like we use fields to keep track of electromagnetic forces, we use **electric potential**, or **voltage** to keep track of electric potential energy. So instead of looking for the potential energy of specific objects, we define it in terms of properties of the space where the objects are.

The **electric potential difference***,* or **voltage difference** (often just called voltage) between two points (A and B) in the presence of an electric field is defined as the work it would take to move a **positive test charge of magnitude 1** from the first point to the second against the electric force provided by the field. For any other charge \begin{align*} q \end{align*}

\begin{align*}\Delta V_{AB} &= \frac{W_{AB}}{q} \ \ \ \ \text{[4] Electric Potential}\end{align*}

Rearranging, we obtain:

\begin{align*}\underbrace{W}_{\text{Work}} &= \underbrace{\Delta V_{AB}}_{\text{Potential Difference}} \times \underbrace{q}_{\text{Charge}}\end{align*}

The potential of electric forces to do work corresponds to electric potential energy:

\begin{align*}\Delta U_{E,AB} &= q\Delta V_{AB} \ \ \text{[5] Potential energy change due to voltage change}\end{align*}

The energy that the object gains or loses when traveling through a potential difference is supplied (or absorbed) by the electric field --- there is nothing else there. Therefore, it follows that electric fields contain energy.

\begin{align*}E = \frac{- \Delta V}{\Delta x} \text{ }\end{align*} Electric field vs electric potential.

\begin{align*}\Delta U_E = q \Delta V \text{ }\end{align*} Change in potential energy due to travel through changing voltage.

\begin{align*}V = \frac{k q}{r} \text{ }\end{align*} Electric potential of a single charge.

To summarize: just as an electric field denotes force per unit charge, so electric potential differences represent potential energy differences per unit charge. Voltage is by definition the electric potential energy per Coulomb. So it is the electrical potential energy value divided by the charge. Thus, voltage difference is the potential value for potential energy. A 12V battery can not produce energy without charge flowing (i.e. you must connect the two ends). Electric potential is measured in units of Volts \begin{align*}(V)\end{align*} – thus electric potential is often referred to as “voltage.” Electric potential is the source of the electric potential energy. You can read the electric potential lines (that is the voltage lines) just like you would a contour map while backpacking in the mountains. Positive charges move towards lower electric potential; negative charges move toward higher electric potential. Thus, positive charges go 'downhill' and negative charges go 'uphill'.

#### Example

You have a negative charge of unknown value and a positive charge of magnitude \begin{align*}q_1\end{align*} and mass \begin{align*}m\end{align*}. After fixing the negative charge in place, you place the positive charge a distance \begin{align*}r_i\end{align*} away from the negative charge and then release it. If the speed of the positive charge when it is a distance \begin{align*}r_f\end{align*} away from the negative charge is \begin{align*}v\end{align*}, what was the magnitude of the negative charge in terms of the given values?

There are multiple ways to do this problem, we will solve it using conservation of energy and the change in voltage to determine the magnitude of the negative charge. When working through this problem, we'll call the positive charge \begin{align*}q_1\end{align*} and the negative charge \begin{align*}q_2\end{align*}. To start we'll say that the charge had zero potential energy when it was .5m from the negative charge; this will help us as we work through the problem. Using this assertion, we will apply conservation of energy to the positive charge.

\begin{align*} \Delta U_e&=KE_f && \text{start with conservation of energy}\\ q_1\Delta V&=\frac{1}{2}mv^2 && \text{substitute the equations for each energy term}\\ \Delta V&=\frac{mv^2}{2q_1} && \text{solve for V}\\ \end{align*}

Now, since we know the voltage difference, we will express it using the equation for voltage at a certain distance from a point charge.

\begin{align*} \Delta V&=\frac{kq_2}{\Delta r} && \text{start with the equation for voltage at a certain distance}\\ \Delta V&=\frac{kq_2}{r_f}-\frac{kq_2}{r_i} && \text{express the change in radius in terms of the initial and final radius of the positive charge}\\ \Delta V&=kq_2(\frac{1}{r_f}-\frac{1}{r_i}) && \text{factor the equation}\\ \frac{mv^2}{2q_1}&=kq_2(\frac{1}{r_f}-\frac{1}{r_i}) && \text{substitute in the value from the first step}\\ q_2&=\frac{mv^2}{2q_1k(\frac{1}{r_f}-\frac{1}{r_i})} && \text{solve for }q_2\\ \end{align*}

### Interactive Simulation

### Review

- The diagram to the right shows a negatively charged electron. Order the electric potential lines from greatest to least.
- \begin{align*}A, B, C\end{align*}
- \begin{align*}C, B, A\end{align*}
- \begin{align*}B, A, C\end{align*}
- \begin{align*}B, C, A\end{align*}
- \begin{align*}A = B = C \ldots\end{align*} they’re all at the same electric potential

- Below are the electric potential lines for a certain arrangement of charges. Draw the direction of the electric field for all the black dots.
- A metal sphere with a net charge of \begin{align*}+5\ \mu \mathrm{C}\end{align*} and a mass of \begin{align*}400 \;\mathrm{g}\end{align*} is placed at the origin and held fixed there.
- Find the electric potential at the coordinate \begin{align*}(6 \;\mathrm{m}, 0)\end{align*}.
- If another metal sphere of \begin{align*}-3 \ \mu \mathrm{C}\end{align*} charge and mass of \begin{align*}20 \;\mathrm{g}\end{align*} is placed at the coordinate \begin{align*}(6 \;\mathrm{m}, 0)\end{align*} and left free to move, what will its speed be just before it collides with the metal sphere at the origin?

- Collisions of electrons with the surface of your television set give rise to the images you see. How are the electrons accelerated to high speed? Consider the following: two metal plates (The right hand one has small holes allow electrons to pass through to the surface of the screen.), separated by \begin{align*}30 \;\mathrm{cm}\end{align*}, have a uniform electric field between them of \begin{align*}400 \;\mathrm{N/C}\end{align*}.
- Find the force on an electron located at a point midway between the plates
- Find the voltage difference between the two plates
- Find the change in electric potential energy of the electron when it travels from the back plate to the front plate
- Find the speed of the electron just before striking the front plate (the screen of your TV)

**Review (Answers)**

- b
- The field lines would point towards the circle, perpendicular to their nearest equipotential.
- a. \begin{align*}7500\mathrm{V}\end{align*} b. \begin{align*}1.5 \;\mathrm{m/s}\end{align*}
- a. \begin{align*}6.4 \times 10^{-17}\;\mathrm{N}\end{align*} b. \begin{align*}1300\mathrm{V}\end{align*} c. \begin{align*}2.1 \times 10 ^{-16} \;\mathrm{J}\end{align*} d. \begin{align*}2.2 \times 10^7 \;\mathrm{m/s}\end{align*}