Coulomb's law 

​​In a nutshell

Electric fields are very similar to gravitational fields. Coulomb's law gives an expression for the force between two charges separated by a distance $$r$$. Electric field strength is the force per unit charge. ​

Equations

Constants

Variable definitions

Electric fields

Electric fields are regions where charged particles experience a force. Electric fields are very similar to gravitational fields except that the fields are to do with charges as opposed to mass.

An electric field can be attractive or repulsive as charges can be positive or negative. Any charged particle or object will have an electric field around it.

Opposite charges attract and similar charges repel. 

Example

A negative charge placed near a positive charge will experience an attractive force, however a negative charge placed near another negative charge will experience a repulsive force.

If the charged object is a sphere then the charge can be assumed to be evenly distributed across the sphere and the maximum field strength is at the surface.  If the object is a point charge or particle, then you can assume all of the charge acts at its centre. 

Electric field strength

Electric field strength is defined as the force per unit charge, similar to gravitational field strength is defined as the force per unit mass. It is given as:

​$$E=\frac FQ$$​​

The field strength depends on the distance to the charge and the strength of the charge. 

Electric field lines

Field lines represent the direction and strength of a field. The closer the lines are together, the stronger the field. 

The field lines are drawn perpendicular to the surface of the charge. For a spherical charge, the field lines will be radial.

The direction of the field is the direction a positive charge would move in the field. If a positive charge was placed near another positive charge, it would move away, therefore the field lines for a positive charge point outwards. 

Coulomb's law

Coulomb's law is the same as Newton's law of gravitation, with the difference being that the masses are switched for charges and the proportionally constant is different. 

Coulomb's law is given as the following (where $$\epsilon_0$$ is the permittivity of free space): 

​$$F=\frac {Qq}{4 \pi \varepsilon_0 r^2}$$​​

Note: The force on $$Q$$ is always equal and opposite to the force on $$q$$ and the direction depends on the charges. 

​​Example

An alpha $$\alpha$$​ particle consisting of two protons and two neutrons is fired towards a gold nucleus $$_{79}^{197}Au$$. When the alpha particle is at a distance of $$160 \, fm$$, the alpha particle is momentarily stationary. Calculate the force between the alpha particle and the gold nucleus.

$$e=1.6\times10^{-19} \, C$$​​

Firstly, write down the known values: 

$$Q_{Gold}=79e \newline \\[0.1in]q_{\alpha} = 2e \newline \\[0.1in]r=160 \, fm = 160 \times 10^{-15} \, m$$

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

$$F=\frac {Qq}{4 \pi \varepsilon_0 r^2}$$

Then, substitute the values into the equation:

$$F=\frac {(79\times 1.6\times10^{-19})(2\times 1.6\times10^{-19})}{4 \pi \varepsilon_0 \times (160\times10^{-15})^2} \newline \\[0.1in]F=1.420705...$$

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

$$electrostatic \ force, F = 1.4 \, N$$​​​​​

The electrostatic force acting between the gold nucleus and the alpha particle is $$\underline{1.4 \, N}.$$​

​

Electric field strength in a radial field

If the electric field is being generated by a point charge, then Coulomb's law can be used in addition with the equation for electric field strength to obtain an equation for the electric field strength in a radial field:

​$$E=\frac Fq \rightarrow E=\frac {\frac {Qq}{4 \pi \varepsilon_0 r^2}}{q} \newline \\[0.1in]E=\frac {Q}{4 \pi \varepsilon_0 r^2}$$

​​

Note: In these equations, $$Q$$ is the charge which is producing the electric field and $$q$$ is the charge experiencing the force from the electric field.  

​