# A-level Physics (Advancing Physics)/Electric Potential Energy

Just as an object at a distance r from a sphere has gravitational potential energy, a charge at a distance r from another charge has electrical potential energy εelec. This is given by the formula:

${\displaystyle \epsilon _{elec}=V_{elec}q}$,

where Velec is the potential difference between the two charges Q and q. In a uniform field, voltage is given by:

${\displaystyle V_{elec}=E_{elec}d}$,

where d is distance, and Eelec is electric field strength. Combining these two formulae, we get:

${\displaystyle \epsilon _{elec}=qE_{elec}d}$

For the field around a point charge, the situation is different. By the same method, we get:

${\displaystyle \epsilon _{elec}={\frac {-kQq}{r}}}$

If a charge loses electric potential energy, it must gain some other sort of energy. You should also note that force is the rate of change of energy with respect to distance, and that, therefore:

${\displaystyle \epsilon _{elec}=\int {F\;dr}}$

## The ElectronvoltEdit

The electronvolt (eV) is a unit of energy equal to the charge of a proton or a positron. Its definition is the kinetic energy gained by an electron which has been accelerated through a potential difference of 1V:

1 eV = 1.6 x 10−19 J

For example: If a proton has an energy of 5MeV then in Joules it will be = 5 x 106 x 1.6 x 10−19 = 8 x 10−13 J.

Using eV is an advantage when high energy particles are involved as in case of particle accelerators.

## Summary of Electric FieldsEdit

You should now know (if you did the electric fields section in the right order) about four attributes of electric fields: force, field strength, potential energy and potential. These can be summarised by the following table:

 Force ${\displaystyle F_{elec}={\frac {-kQq}{r^{2}}}}$ → integrate → with respect to r Potential Energy ${\displaystyle \epsilon _{elec}={\frac {-kQq}{r}}}$ ↓ per. unit charge ↓ Field Strength ${\displaystyle E_{elec}={\frac {-kQ}{r^{2}}}}$ → integrate → with respect to r Potential ${\displaystyle V_{elec}={\frac {-kQ}{r}}}$

This table is very similar to that for gravitational fields. The only difference is that field strength and potential are per. unit charge, instead of per. unit mass. This means that field strength is not the same as acceleration. Remember that integrate means 'find the area under the graph' and differentiate (the reverse process) means 'find the gradient of the graph'.

## QuestionsEdit

k = 8.99 x 109 Nm2C−2

1. Convert 5 x 10−13 J to MeV.

2. Convert 0.9 GeV to J.

3. What is the potential energy of an electron at the negatively charged plate of a uniform electric field when the potential difference between the two plates is 100V?

4. What is the potential energy of a 2C charge 2 cm from a 0.5C charge?

5. What is represented by the gradient of a graph of electric potential energy against distance from some charge?

Worked Solutions