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(1)

B.Tech Physics Course NIT Jalandhar

electrostatics Lecture 3

Dr. Arvind Kumar Physics Department

(2)

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)

(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,

(4)

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

(5)

Also the P.D. between two points can be written as

Also using fundamental theorem for gradients, we have

(6)

Like electric force and electric field, electric potential also obeys the superposition principle

(7)

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

(8)

In general the potential for a positive point charge is

For a collection of charges, we have

(9)

For volume charge distribution we have

(10)

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)

(11)

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

(12)

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)

(13)

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

(14)

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

(15)

We can also write

We can write

(16)

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,

(17)

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 .

(18)

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,

(19)

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

(20)

Using Legendre Polynomial Eq. (6) can be written as

---(7)

Using Eq. (7) in Eq. (1) we get

---(8)

Or we write

(21)

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

(22)

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)

(23)

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

References

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