Book Reference : Pages & To look at the work of Charles Coulomb 2.To understand electric potential

Around 1784 Coulomb devised an experiment to establish the force between charged object using a torsion balance 1.Like charges repel 2.A pair of charged pith balls 3.Suspended from a vertical wire 4.2 nd charged ball 5.Wire twisted until it balanced the repulsion 6.Distance between balls varied

Here are some of Coulomb’s results can you spot a pattern between F and r? Distance, r Force, F Note both results were measured in degrees and so the units are relative

Halving the distance from 36 to 18 makes the force increase by x4 and again from 18 to 8.5 F  1/r 2 The force is also proportional to the size of the two forces involved F  Q 1 Q 2

Bringing this together F  Q 1 Q 2 r 2 As usual we can turn a proportionality into an equation by introducing a suitable constant of proportionality F = 1Q 1 Q 2 4  0 r 2

Definition : The magnitude of the force F between two electrically charged bodies, which are small compared to their separation r is inversely proportional to r 2 and proportional to the product of their charges Q 1 and Q 2 Compare the form of Coulomb’s law with Newton’s law of gravitation

Not required for A2 Permittivity describes how an electric field affects, and is affected by, a dielectric medium, and is determined by the ability of a material to polarize in response to the field, and thereby reduce the total electric field inside the material. Thus, permittivity relates to a material's ability to transmit (or "permit") an electric field. Permeability is a constant of proportionality that exists between magnetic flux density and magnetic field strength in a given medium Experimentally it can be shown that c = 1 /  0  0

Calculate the force between an electron and a.A proton at a distance of 2.5x10 -9 m b.The nucleus of a nitrogen atom (atomic number 7) at a distance of 2.5x10 -9 m [3.7 x N] [2.6 x N] e = -1.6 x C  0 = 8.85 x F/m

Two point charges Q 1 is +6.3nC & Q 2 is - 2.7nC exerts a force of 3.2x10 -5 N when they are d metres apart a.Find d b.Find the force if d increases to 3d [69mm] [3.6 x N] e = -1.6 x C  0 = 8.85 x F/m

We know like charges repel. To bring two like charges X & Y together, work must be done. The field around Y must be overcome As we move X from  towards Y the electric potential energy of X increases from 0 Definition : The electric potential at a certain point in any electric field is defined as the work done per unit charge on a “small +ve test charge” when it is moved from  to that point

For a +ve test charge in a field where the electric potential energy is E p, the electric potential V is shown by: V = E p / Q Where Q is the charge in Coulombs, E p is the electric potential energy (J) & V is the electric potential in Volts or J/C

If test charge Q is moved from a point in the field where the potential is V 1 to a point where the potential is V 2, then the work done  W is given by  W = Q(V 2 –V 1 )

Example : If a + 1  C test charge is moved into an electric field from  to a point where the electric potential is 1000V. The electric potential energy is given by E p = QV E p = 1 x x 1000 = 1x10 -3 J i.e. 1x10 -3 J of work has been done moving the charge from infinity to P