Electric potential and potential energy

In a nutshell

Electric potential is the work done per unit of charge whereby the charge is positive and moved from infinity to a point in an electric field. The value of electric potential is negative if the charge is negative and the force will be attractive. The value of electric potential is positive if the charge is positive and the force will be repulsive.

Equations

Description	Equation
Electric potential	$V=\frac Q{4\pi \varepsilon_0 r}$
Electric potential energy	$energy=\frac {Qq}{4\pi \varepsilon_0 r}$
Capacitance of charged sphere	$C=4\pi \varepsilon_0 R$

Constants

name	symbol	value
$permittivity \ of \ free \ space$	$\varepsilon_0$	$8.85 \times 10^{-12} \, F \, m^{-1}$

Variable definitions

Quantity Name	Symbol	Derived Unit	SI BASE Units
$electric \ potential$	$V$	$V$	$kg\,m^2\,s^{-3}\, A^{-1}$
$charge \ creating \ e-field$	$Q$	$C$	$A\,s$
$charge \ in \ e-field$	$q$	$C$	$A\,s$
$energy$	$energy$	$J$ $eV$	$kg\,m^2\,s^{-2}$
$distance$	$r$	$m$	$m$
$capacitance$	$C$	$F$	$kg^{-1}\,m^{-2}\,s^4\,A^2$
$radius\ of\ sphere$	$R$	$m$	$m$

Electric potential

Electric potential is the work done per unit of charge whereby the charge is positive and moved from infinity to a point in an electric field. At infinity the electric potential would be zero.

In a radial field around a point charge or charged sphere, the electric potential can be calculated using:

$V=\frac Q{4\pi \varepsilon_0 r}$

Curiosity: You may have heard the term potential difference when learning about circuits. This is the same as electric potential difference as they are both defined as the work done per unit charge.

Electric potential can be positive or negative depending on the charge creating the electric field.

If the charge is positive, then electric potential will be positive. If the charge is negative, then the electric potential will be negative.

Graphically this looks like this:

Physics; Electric fields; KS5 Year 12; Electric potential and potential energy

1	Electric potential
2	Positive charge
3	Negative charge
4	Distance

Note: The absolute magnitude of electric potential is the greatest on the surface of the charge.

Electric potential energy

Electrical potential energy is equal to the work done done bringing a particle from infinity to a distance of $r$ .

Electric potential energy is directly proportional to electric potential.

Example

A charge of $+5\, C$ would mean the work done would be $5$ times greater than a charge of $+1 \ C$ .

Turning this into a universal equation for any charge:

$energy=Vq$

Note: $q$ in this equation is the charge being moved into the electric field being created by $Q$ .

Substituting in electric potential into the equation for energy:

$energy=(\frac Q{4\pi \varepsilon_0 r})q \newline \\[0.1in]energy=\frac {Qq}{4\pi \varepsilon_0 r}$

When a charge is moved in an electrical field, there must be a force which is doing the work. Using Coulomb's law, the force is inversely proportional to the distance squared, $F \propto \frac 1{r^2}$ .

For a radial field, the graph for moving a negative charge within an electrical field is shown below:

Physics; Electric fields; KS5 Year 12; Electric potential and potential energy

Note: The graph for a positive charge in an electric field will be the same just reflected in the $x$ -axis.

The area under the curve between two different distances is equal to the work done moving the charge from $R_1$ to $R_2$ .

Capacitance of a charged sphere

Using the equation for electric potential, it is possible to derive an expression for the capacitance of a charged sphere.

Recall the equation for capacitance:

$C=\frac QV \rightarrow V=\frac QC$

Then substitute in the equation for electric potential:

$\frac QC=\frac {Q}{4\pi \varepsilon_0 R}$

Note: $R$ in this equation is the radius of the charged sphere and not the distance away from the charge. This is because the potential, in this instance, is at the surface of the sphere.

Simplify through and rearrange for capacitance:

$\frac 1C=\frac {1}{4\pi \varepsilon_0 R} \newline \\[0.1in]C=4\pi \varepsilon_0 R$

Example

What is the electric potential of a uranium $_{92}^{235}U$ nucleus at a distance of $3.0 \,fm$ from the centre?

Firstly, write down the known values:

$Q_{U}=92 \times 1.6\times10^{19} \, C \newline \\[0.1in]r=3.0\, fm=3.0\times 10^{-15} \, m$

Next, write down the equations needed and rearrange if necessary:

$V=\frac Q{4\pi \varepsilon_0 r}$

Then, substitute the values into the equation:

$V=\frac {92\times 1.6\times 10^{-19}}{4\pi \times 8.85 \times 10^{-12} \times 3.0\times 10^{-15}} \newline \\[0.1in]V=44\,119\,788... \, V$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$electrical\, potential, \ V=44\, 000\,000 \, V$

The electric potential $3\, fm$ away from a uranium nucleus is $\underline{44\, 000\,000 \, V}$ , which is equivalent to $\underline{44 \ M V}$ .

Electrical and gravitational fields

There are many similarities between electrical and gravitational fields. Newton's law and Coulomb's law are nearly identical aside from the constant of proportionality and the fact one is masses and the other is charges.

They both have inverse square laws for the forces and their potentials are inversely proportional to distance.

However there are a few key differences about electric and gravitational fields:

Gravitational fields are always attractive as you can't get a negative mass.
Electric fields can be shielded against, gravitational fields can not.
The medium charges are in, affects the electric field, whereas the medium is negligible for gravitational fields.