B.Tech Physics Course NIT Jalandhar
electrostatics Lecture 3
Dr. Arvind Kumar Physics Department
The curl of electric field: The electric field due to a point charge is given by
---(1)
Now we calculate the line integral of the electric field from some point a to b i.e. We shall find
---(2) In spherical co-ordinates
---(3)
So now find the dot product of E and dl, using (2) and (3)
Using Eq. (4) in (2), we get
- ---(5)
For closed path, the end points coincides, so we have
Using Stokes theorem above eq. Gives us,
Electric Potential:
We know
Above eqn implies that the line integral of E around any closed path zero
This means line integral from point a to point b is same for
different paths. If it is not so then one can go out along path (1)
and return through path (ii) and But this is not the Case.
Thus we define a function
Also the P.D. between two points can be written as
Also using fundamental theorem for gradients, we have
Like electric force and electric field, electric potential also obeys the superposition principle
The potential of a localized charge distribution:
Given a charge distribution, we shall find the electric potential and
then taking the gradient of this we shall find the electric field
Taking the reference point at infinity, the potential of a point charge is given as
In general the potential for a positive point charge is
For a collection of charges, we have
For volume charge distribution we have
Work and energy in electrostatics:
Work done to move a point charge:
Consider a stationary charge distribution.
We want to find the work done to move the point charge Q from point a to point b.
Force on charge Q, F = QE
To move the charge we need to exert the force in opposition to above force i.e. the force –QE
Work done,
---(1)
Eqn (2) shows that the work done is independent of path followed in going from a to b. Such electrostatic force is known as
conservative force.
From eqn (2) we can write
Above eqn defines the potential difference between two points as the work done per unit charge to carry particle from point a to b
If we move the charge from infinite to some point r then
The energy of a point charge distribution:
Here we shall try to find the work required to be done to assemble the point charges. The work done to move the first point charge to point r1 is zero as there is no field against which we need to fight i.e. W1 = 0
Now we want to find the work done to move the 2nd charge. Here
we have to fight against the field of charge q1. Thus we have ---(1)
Now when we move the 3rd charge i.e. the charge q
3 to its position
then we need to fight against the field of q1 and q2
---(3)
Similarly the work done to assemble the 4th charge is
Total work done to assemble the first four charges is given as
---(6)
In general , take the product of each pair of charge and add them
We can also write
We can write
Electric potential due to a dipole: An electric dipole consists of two equal and opposite charges say +q and –q separated by distance d.
Let the observation point is at distance from the +q charge and at distance from –q charge.
Then using the principle of superposition we write the total electric potential at P as,
---(1) Using cosines law we can write the following Eqn. ,
---(2) In the region where , the third tem in above expression is
negligible, So we write using binomial expansion,
Using Eq. (3) we write
---(4)
Using Eq. (4) in Eq. (1) we get
---(5)
Above Eq. gives us the electric potential due to dipole at large distance. We observe that potential goes like .
Multiple expansion: We shall consider a localized charge distribution and shall find the electric potential due to this at some far off
distance r .
In general the electric potential is given by
---(1)
Where is the small volume element of the charge distribution and ρ is the volume charge
density.
Using the cosines law we write
---(2)
Which we write further as
---(3)
Where,
For points which are at very large distance from the charge distribution we use the binomial expansion and write
---(5) Using (4) in (5), we get
Using Legendre Polynomial Eq. (6) can be written as
---(7)
Using Eq. (7) in Eq. (1) we get
---(8)
Or we write
Eq. (8) or (9) gives us the multipole expansion of the potential V.
The first term corresponding to n = 0, gives the contribution from the monopole and it goes like 1/r. The 2nd term gives the
Monopole and Dipole terms in multipole expansion: The multipole expansion is normally dominated by the monopole term, which is
written as
---(1)
Where is the total charge of the configuration.
If the total charge of the configuration is zero then the potential will be dominated by the dipole configuration and is given as,
---(2) Since is the angle between and r
---(3)
Using Eq. (3) in Eq. (2), we get
---(4)
In above Eq. term with integral is known as dipole moment of the Configuration
---(5)
Using (5) in (4), we get