PHYS 172

Lecture 20a - Electric Potential

Introduction

Electric potential is more commonly known as voltage. It is produced by all electric charges, but unlike electric field, electric potential is a scalar (as opposed to a vector). The symbol for electric potential is $V$, but confusingly we use the same symbol $V$ to represent the SI unit, volts. Voltage could be positive or negative, for example: $$ \begin{eqnarray} V_1 &=& +5V\\ V_2 &=& -7V \end{eqnarray} $$

Electric Potential from Point Charges

The electric potential at distance $r$ from a point charge $q$ is given by:

$$ \begin{eqnarray} V &=&\frac{q}{4\pi \epsilon_0 r} \end{eqnarray} $$

Just like before, we often use $k = \frac{1}{4\pi \epsilon_0} \approx 8.99\times 10^9 N m^2 C^{-2}$ to write the above equation as: $$ \begin{eqnarray} V &=&\frac{k q}{r} \end{eqnarray} $$

In some problems below, we will often refer to a point at infinity, meaing a point very far away from all charges. It is the same as setting $r\rightarrow \infty$, which gives: $$ V_\infty = 0V $$

Simulation - Electric Potential of Two Charges (click to hide)

Canvas not supported

Electric Potential of Two Charges

The charges and the observer can all be moved.
The charge can be changed by selecting it and then adjusting the slider.
Calculations will appear here.
Question will be loaded by load_exercise_example_all() defined in script_question.js
Question will be loaded by load_exercise_example_all() defined in script_question.js

Try It Yourself (click to show)

Question will be loaded by load_exercise_example_all() defined in script_question.js
Question will be loaded by load_exercise_example_all() defined in script_question.js

Potential Difference

Potential difference is defined as the difference between the electric potential of two points. For example, if an observer travels from point $i$ to point $f$, the potential difference of this journey is given by: $$ \Delta V_{fi} = V_f - V_i $$ Recall that the symbol $\Delta$ means change, and is always computed by "final minus initial" (the order is important!).

For example, if you used to be have $V_i = 10V$ at point $i$, then you moved to point $f$ where $V_f = 80V$, then the change in electric potential (i.e. potential difference) you experienced is $\Delta V_{fi} = 80V - 10V = +70V$.

Note that by this definition, we have: $$ \Delta V_{AB} = - \Delta V_{BA} $$ because $V_A - V_B = -(V_B - V_A) $.

Try It Yourself (click to show)

Question will be loaded by load_exercise_example_all() defined in script_question.js
Question will be loaded by load_exercise_example_all() defined in script_question.js
Question will be loaded by load_exercise_example_all() defined in script_question.js

Work Done by the Electric Field

When a charge moves through an electric field, there is an exchange of energy between the field and the charge. In physics, the transfer of energy is called work (SI unit: $J$). The amount of energy transferred from the electric field to a charge is the "work done by the electric field", $W_E$. For example, if $W_E = 100J$, it means the electric field has just transferred $100J$ to the charge.

$W_E$ for a charge moving from point $i$ to point $f$ can be calculated using the potential difference $\Delta V_{fi} = V_f - V_i$:

$$ W_E = - q \Delta V_{fi} $$

The Work Kinetic Energy Theorem states $W_{total} = \Delta KE$. In the special case (which is usually the case we are interested in below) when the electric force is the only force acting on the charge, we have: $$ \begin{eqnarray} W_E &=& \Delta KE = KE_f - KE_i \\ \Rightarrow KE_f &=& KE_i + W_E \end{eqnarray} $$ We see that the energy transferred from the electric field simply becomes the kinetic energy of the charge.

Question will be loaded by load_exercise_example_all() defined in script_question.js

Try It Yourself (click to show)

Question will be loaded by load_exercise_example_all() defined in script_question.js
Question will be loaded by load_exercise_example_all() defined in script_question.js
Question will be loaded by load_exercise_example_all() defined in script_question.js

Potential Energy

A pair of charges $q_1$ and $q_2$ separated by $r_{12}$ carries an amoung of potential energy given by:

$$ U_{12} = \frac{q_1 q_2}{4\pi \epsilon_0 r_{12}} $$
This SI unit for $U$ is $J$, the same as the unit of energy. Potential energy could be positive (when the charges have the same signs) or negative (when the charges have the opposite signs).

Do not confuse potential energy $U$ with potential $V$. They do not even have the same unit ($J$ vs $V$)!

Question will be loaded by load_exercise_example_all() defined in script_question.js

Simulation - Electric Potential Energy (click to hide)

Canvas not supported

Electric Potential Energy

The charges and the observer can all be moved.
The charge can be changed by selecting it and then adjusting the slider.
Calculations will appear here.

Try It Yourself (click to show)

Question will be loaded by load_exercise_example_all() defined in script_question.js

Equipotential Surface

An equipotential surface is a surface on which the electric potential has the same value.

For example, if you are on an equipotential surface and you measure the potential at that point to be $10V$. As long as you stay on the same equpotential surface, the potential will remain at $10V$.

An equipontial surfaces are sometimes called "equipotential lines", or simply "equipotentials".

Electric field lines and equipotentials are always perpendicular to each other whenever they cross.

Simulation - Equipotential (click to hide)

Canvas not supported
You can drag more charges and observers out from the two square boxes at the top right hand corner. Drag them back to remove.
To add equipotential, first select an observer then click the "Add equipotential" button.
Tip: Try not to show both field lines and equipotentials at the same time otherwise it may look too messy. You can control this by using the show/hide buttons.
Can you see the electric field always points perpendicular to the equipotentials?

The Relationship Between Electric Potential and Electric Field

The electric field and electric potential are actually closely related to each other. In one-dimension, the relationship is given by: $$ \begin{eqnarray} E_x &=& -\frac{dV}{dx} \end{eqnarray} $$ Roughly speaking, $E_x$ measures how rapidly $V$ is changing in the $x$-direction, and $\vec{E}$ always points in the direction of decreasing $V$. For example, when $V$ is a constant, then $E_x=0V/m$. If for every meter you move to the right, the potential drops by $10V$, then $E_x = -10V/m$.

Since $\vec{E}$ always points in the direction of decreasing $V$, using $\vec{F}_E = q\vec{E}$, one can show that:

Notations

Name Symbol Unit Meaning
Electric Potential $V$ $V$ useful in computing energy
Potential Difference $\Delta V$ $V$ change in potential when moving from one location to another
Work Done by E Field $W_E$ $ J $ energy transferred from E field
Potential Energy $U$ $J$ potential energy stored between pairs of charges