Potential Energy

Conservative Force

The work done on an object by a conservative force depends only on the object's initial and final position, and not the path taken.
The net work done by a conservative force in moving an object around a closed path is zero.

Potential Energy

The electric force, like gravity, is a conservative force.

Let's place a positive point charge $q$ in a uniform electric field and let it move from the positive plate to the negative plate. How much work is done by the field in moving the charge?

*Remember, $W=F*d$ , where $F$ is the component of the constant force along the direction of the motion. Here, $F=qE$ , so $W_{AB}=qE(y_f-y_o)=qE\Delta y=\Delta EPE$ .

The work done = change in electrostatic potential energy!

$\dfrac{W_{AB}}{q}=\boxed{\dfrac{\Delta EPE}{q}}$ The quantity on the right is the potential energy per unit charge. We call this the electric potential, $V$ . $\boxed{V=\dfrac{EPE}{q}}$ - units - $\left[\dfrac{\text{Energy}}{\text{Charge}}\right]= \left[\dfrac{J}{C}\right]=[V \text{ (volts)}]$

Work

Work is not a vector, but it can be either positive or negative:
- Positive - Force is in the same direction as the motion.
- Negative - Force is in the opposite direction as the motion.
If positive work is done on an object, the object speeds up.
If negative work is done on an object, the object slows down.

Electric Potential Difference

We can determine the value of the potential at different points in space. For example, what is the difference in electrostatic potential between two points, A and B, in an electric field? electric potential $V_B-V_A=\dfrac{EPE_B}{q}-\dfrac{EPE_A}{q}=\dfrac{-W_{AB}}{q}$ , so $\boxed{\Delta V=V_B-V_A=\dfrac{-W_{AB}}{q}}$ .

Let's say the charge at point A is positive. If I release it, it'll move down towards B. Since the force is down and the motion is down, positive work is done on the charge, so $W_{AB}$ is positive. This means $V_B-V_A$ is negative, or $V_A>V_B$ . Point A is at a higher potential than point B.

Positive charges, starting from rest, will move away from high potential regions and move toward low potential regions.
Negative charges, starting from rest, will move away from low potential regions and move toward high potential regions.

Voltage

$1.5V=1.5\dfrac{J}{C} \rightarrow$ The battery supplies 1.5 Joules of energy for every Coulomb of charge.

Energy

If we accelerate an electron form rest through a potential difference of 1 Volt, then it would gain 1 electron volt ( $eV$ ) of kinetic energy. Energy is usually expressed in Joules: $\boxed{1eV=1.602*10^{-19}J}$

Energy Conversion

Just like in a gravitational field, in an electric field, potential energy ( $PE$ ) can be converted into kinetic energy ( $KE$ ). For example, if we bring a positive charge close to a fixed, positive point charge, the repulsive force on it gets bigger and bigger. We have to do work on the charge to move it closer, which increases its $PE$ . When we release the charge, the stored electric potential energy ( $EPE$ ) is converted into $KE$ !

The total mechanical energy of a system must be conserved.

$E_{\text{Tot}}=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2+mgh+\frac{1}{2}kx^2+EPE$

If the work done by non-conservative forces is zero, then $\boxed{E_{\text{Tot}_f}=E_{\text{Tot}_o}}$ .

Types of Kinetic Energy

$\frac{1}{2} mv^2$ : linear or translational
$\frac{1}{2}I\omega^2$ : rotational
$mgh$ : gravitational
$\frac{1}{2}kx^2$ : elastic
$EPE$ : electrostatic

Electric Potential of a Point Charge

An electric potential exists around charges. What is the form of the potential for a point charge?

If we place a positive test charge near a positive fixed point charge, the electric field lines created by the point charge does work on the test charge and moves it to the right.

The work done on the field when moving the charge is $W_{AB}=k\dfrac{qQ}{r_A}-k\dfrac{qQ}{r_B}$ . Electric potential due to a point charge: $V=k\dfrac{Q}{r}$ .

Equipotential Surfaces

The electric potential of a point charge is $V=k\dfrac{Q}{r}$ . This means the potential is the same in every direction around the point charge at a distance $r$ away. This forms a spherical shell of radius $r$ around the charge.

Thus, the electric potential is the same everywhere on this spherical surface ( $S_A$ ) - called an equipotential surface.

The electric field lines are perpendicular to the equipotential surface.

In another equipotential surface which is further away but still around the point charge ( $S_B$ ), there is a lower potential than there is in $S_A$ . Thus, electric field lines point in the direction of decreasing potential (high --> low potential). Therefore, positive charges move in the same direction as the electric field points, and negative charges move in the opposite direction of the field.

Work Done on Equipotential Surfaces

The net electric force does no work as a charge moves on a equipotential surface. Why? $V_B-V_A=\dfrac{-W_{AB}}{q}$ . But if we're on an equipotential surface, then $V_A=V_B$ , and $W_{AB}=0$ .

(In other words, in order for the charge to feel a force along an equipotential surface, there must be a component of the field along the surface, but $E$ is perpendicular to the surface everywhere.)

Parallel Plate Capacitor

The positive plate is at a potential of $+9\ V$ and the negative plate is at $0\ V$ . The equipotential surfaces between the planes look like a parallel set of plates.

Let the plates be separated by a distance $\Delta s$ . The electric field is then $\dfrac{[\text{Change in voltage}]}{[\text{Change in distance}]}\rightarrow E=\dfrac{\Delta V}{\Delta s}$ (units: $\dfrac{V}{m}$ ). This is called the electric field gradient.

Capacitor

We have two oppositely charged conducting plate separated by some small distance. We charge the plates by connecting them to a battery. The higher the voltage on our battery, the more charge we can put on each plate. Thus, $Q \varpropto V$ .

$\boxed{Q=CV}$ . $C$ is the capacitance. $C=\dfrac{Q}{V} \rightarrow \dfrac{[\text{Charge}]}{[\text{Voltage}]}=\left[ \dfrac{C}{V} \right]=[\text{Farad}]=[F]$

*A farad is a very large capacitance. We often use microfarads ( $\mu f$ : $1*10^{-6}\ F$ ) and picofarads ( $pf$ : $1*10^{-12}\ F$ ).

The larger the capacitance, the more charge it will hold!

Dielectrics

We can fill the space between the plates with some insulating material, say air, oil, paper, rubber, plastic, etc. This material is called a dielectric.

Since the dielectric is an insulator, the charges in it aren't free to move, but they can separate slightly within each atom. Each one of these atoms now produces a small internal electric fields which points in the opposite direction to the field between the plates.

Thus, the net electric field between the plates is reduced by the dielectric.

The reduction of the field: $\boxed{\kappa=\dfrac{E_o}{E}}$ .

$E_o=$ field without the dielectric
$E=$ field with the dielectric
$\kappa=$ dielectric constant (must be greater than 1, unitless)

material	$\kappa$
vacuum	1
air	1.00054
water	80.4

The larger $\kappa$ is, the more it reduces the field between the plates.

Let's say the plates have a surface area $A$ and are separated by a distance $d$ . $E=\dfrac{1}{\kappa}E_o=\dfrac{V}{d} \rightarrow E_o=\dfrac{\kappa V}{d}=\dfrac{\sigma}{\epsilon_o}=\dfrac{q}{\epsilon_o A} \rightarrow \boxed{q=\left( \dfrac{\epsilon_o A \kappa}{d}V \right)}$ , but $q=CV$ , so $\boxed{C=\dfrac{\epsilon_o A \kappa}{d}}$ .

Capacitors store charge. What about energy?

$EPE_{\text{Stored}}=\frac{1}{2}qV=\frac{1}{2}CV^2$

$\boxed{\dfrac{EPE}{\text{Vol}}=\text{Energy Density}=\frac{1}{2}\kappa \epsilon_o E^2}$

units: $\left[ \dfrac{\text{Energy}}{\text{Volume}} \right]=\left[ \dfrac{J}{m^3} \right]$

This expression holds true for any electric fields, not just for capacitors.