110PHY111 Revision Notes — Electrostatics
110PHY111 Revision Notes — Electrostatics
Page 1 of 3
1.
Coulomb’s Law.
F =
1
4 π
0 q
1 q
2 r 2
(N)
The hard thing here is that force is a vector. You want to get the force in the form
(magnitude
× unit vector) see the solution to question 1, Tutorial Question Sheet 6. If necessary deal with the x , y and z components one at a time.
2.
Principle of Superposition.
The force between charges is not affected by the presence of other charges.
Hence the force on each charge is the sum of the forces due to all others calculated one by one using Coulomb’s Law.
3.
Electrostatics.
The whole of electrostatics (indeed the whole of electrodynamics if you admit also the principles of special relativity) rests solely on the last two postulates.
However, given a set of charges or a charge distribution, calculating the force on another charge can be very difficult even in simple geometries as you have seen in class and in Tutorial Question Sheet 6. Therefore we have to learn about electric field, electric potential and their properties. Also Gauss’s Law can be very useful in solving problems of highly symmetric charge distributions.
4.
Electric Field.
This is the force per unit charge. For a general distribution of charge
ρ ( r
1
) C m −
3 , the electric field is
E ( r
0
) =
1
4 π
0
Z
V
ρ ( r
1
) r 2
ˆ d τ
1
(N C
−
1
) r
0 is the field point and r
1 is a source point at the centre of an elemental volume d τ
1 and r is the vector joining the two. The integral is over the volume
V throughout which the density exists. It’s this geometry that make this difficult; look at the figure in your notes at the beginning of section 4.
We also deal with surface distributions of charge having density σ ( r
1
) C m −
2 densities of λ ( r
1
) C m −
1 and linear
5.
Electric Flux.
If the electric field is uniform through a certain surface ( ie , the density of field lines is constant) and penetrates the surface at right angles then the electric flux is electric field times area . Revise the rules for drawing field lines and how the flux is defined most generally if the field is non uniform and the surface you are interested in has arbitrary shape.
6.
Gauss’s Law.
This is important for two reasons. ( i ) It is the first of four Maxwell equations that you have to learn. ( ii ) It can make the calculation of electric field very easy in some circumstances. Note how in Tutorial Question Sheet 8 you were able to do the problem of a long line of charge and a large sheet of charge much more easily than the way it was done using Coulomb’s Law in class.

110PHY111 Revision Notes — Electrostatics Page 2 of 3
Gauss’s Law states that the total flux penetrating a closed surface is the charge enclosed divided by
0
. That is,
I
E · d a =
1
0
Q enclosed
It is simply a consequence of Coulomb’s Law and the principle of superposition and is true because of the inverse square nature of the force. (There is also a “Gauss’s Law” in gravitation.) The differential form of Gauss’s Law is
·
E =
1
0
ρ
Here is a vector (see the Mathematical notes )
∂
∂x
∂
∂y
+
∂
∂z
7.
Curl of E.
The electric field is irrotational: there is no circulation in the electric field.
This is true as long as all charges are stationary. This is expressed mathematically as a line integral along a closed loop,
I
E · d l = 0
The differential form of this is
× E = 0 where this is the vector cross product between and E (see the Mathematical notes ).
8.
Electric Potential.
This is the work done in moving a unit charge from some agreed place to a field point r
0 in an electric field. If this agreed place is at infinity ( ie , away from the influence of any charge) then the electric potential is
V ( r
0
) =
1
4 π
0
Z
V
ρ ( r
1
) r d τ
1
(V)
One volt (V) is one N m C −
1 or one J C −
1 . Do not confuse electric potential with potential energy. A point charge q has a potential energy (with respect to infinity) of q V ( r ) (J) when placed at r where the electric potential is V ( r ). The electric potential is the potential energy per unit charge, in analogy with the electric field which is the force per unit charge.
The potential difference between two points b and a in an electric field is the difference in electric potentials at the two points. It is also the work done on a unit charge in bringing it from point a to point b ,
V ( b )
−
V ( a ) =
−
Z b a
E · d l
It is independent of the path taken in getting from a to b .
The relation between electric potential and electric field is
E =
−
V

110PHY111 Revision Notes — Electrostatics Page 3 of 3
9.
Electrostatic energy.
A charge distribution ρ ( r
1
) C m −
3 has an associated self energy , which is the work done in assembling the distribution from infinitessimally small increments of charge taken from infinity. This energy is
W =
=
1
2
1
2
Z
ρ V d τ
0
V Z
E
2 d all space
τ where the first integral is over the entire volume within which the charge density is not zero and the second integral is over all space (assuming no other fields exist due to other charges).
10.
Examples in electrostatics.
You should know the electric fields due to
1. A point charge.
2. A dipole (at least parallel and perpendicular to the centre).
3. A line of uniform charge density λ ( r
0
) C m −
1 .
4. A sheet of uniform charge density σ ( r
0
) C m −
2 .
You should know that these particular fields fall off with distance like
1. 1 /r
2
2. 1 /r
3
3. 1 /r
4. not at all.
In cases 1, 3 and 4 you should know how to obtain the electric field using Gauss’s Law.
Why can’t you use Gauss’s Law in the case of the dipole, case 2?