Transcript L24_A2_2009_10_CoulombsLaw

Book Reference : Pages 83-85 & 80-81
1.
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.
2.
3.
4.
5.
6.
Like charges repel
A pair of charged pith
balls
Suspended from a vertical
wire
2nd charged ball
Wire twisted until it
balanced the repulsion
Distance between balls
varied
Here are some of Coulomb’s results can
you spot a pattern between F and r?
Distance, r 36
Force, F
36
18
144
8.5
567
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/r2
The force is also proportional to the size of
the two forces involved
F  Q1Q2
Bringing this together
F  Q1Q2
r2
As usual we can turn a proportionality into
an equation by introducing a suitable
constant of proportionality
F = 1 Q1Q2
4 r2
0
Definition :
The magnitude of the force F between
two electrically charged bodies, which
are small compared to their separation r
is inversely proportional to r2 and
proportional to the product of their
charges Q1 and Q2
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 2.5x10-9m
b. The nucleus of a nitrogen atom (atomic
number 7) at a distance of 2.5x10-9m
[3.7 x 10-11N] [2.6 x 10-10N]
e = -1.6 x 10-19 C
0 = 8.85 x 10-12 F/m
Two point charges Q1 is +6.3nC & Q2 is 2.7nC exerts a force of 3.2x10-5N when they
are d metres apart
a. Find d
b. Find the force if d increases to 3d
[69mm] [3.6 x 10-6N]
e = -1.6 x 10-19 C
0 = 8.85 x 10-12 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 Ep, the electric
potential V is shown by:
V = Ep / Q
Where Q is the charge in Coulombs, Ep 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 V1 to a point
where the potential is V2, then the work
done W is given by
W = Q(V2 –V1)
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
Ep = QV
Ep = 1 x 10-6 x 1000 = 1x10-3 J
i.e. 1x10-3J of work has been done moving the
charge from infinity to P