B.
Tech Physics Course NIT Jalandhar
electrostatics Lecture 3
Dr. Arvind Kumar
Physics Department
e.mail. : iitd.arvind@gmail.com
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)
---------(4)
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
where O is some reference point. The value of above function depend
upon point r only. Above function is know as electric potential.
Also the P.D. between two points can be written as
Also using fundamental theorem for gradients, we have
Above eqn show that the electrostatic field can be written as the
-ve of the gradient of the scalar electric potential
Like electric force and electric field, electric potential also obeys
the superposition principle
Units: Joule per Coulomb or volt
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
Note the potential of a positive charge is positive (hills)and the
potential of a negative charge is negative (valleys) and electric
field point from +Ve to Ve direction
In general the potential for a positive point charge is
For a collection of charges, we have
For a continuous charge distribution
For volume charge distribution we have
For line and surface charge 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
-----------------(1)
To move the charge we need to exert the force in opposition to
above force i.e. the force QE
Work done,
-----------------(2)
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
Thus the potential is the potential energy (work done to create the
system) per unit charge.
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
---------------(1)
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
---------------(2)
Now when we move the 3rd charge i.e. the charge q3 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
---------------(5)
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
j>i tells us not to count each pair twice.
We can also write
We can write
Term in brackets represents the potential at ri (position of qi)
due to all other charges
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,
--------(3)
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
.
Note the following figure:
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
is the volume charge
density.
Using the cosines law we write
------------(2)
Which we write further as
-------------(3)
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
Where we have collected the like powers of
Are the Legendre polynomials.
------------(6)
and the coefficients
Using Legendre Polynomial Eq. (6) can be written as
--------(7)
Using Eq. (7) in Eq. (1) we get
------------(8)
Or we write
----------(9)
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
contribution from a dipole and it goes like 1/r2 .
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 r 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
------------(6)
