Electric Dipole , Dipole Moment & Electric Field Intensity due to Dipole

Electric Dipole :

A combination of equal and opposite charges kept close to each other is called an electric Dipole. It is characterised by dipole moment vector .

The following results can be established for a dipole.

 Dipole Moment :

Dipole moment is a vector quantity whose magnitude is equal to the product of the magnitude of charges forming the dipole and their separation.

Dipole Moment , $\large \vec{p} = (q \times 2a) \hat{i}$

Dipole moment vector is directed from the negative to the positive charge.

 Electric field Intensity due to an electric dipole

At a Point on axial Line :  We have to find Electric field intensity at Point P .

Consider AB is an electric dipole having mid point O & length 2a  .

Let OP = r

Electric field  Intensity at a Point P due to A(-q) is

$\large E_{PA} = \frac{1}{4\pi \epsilon_0} \frac{q}{AP^2}$

$\large E_{PA} = \frac{1}{4\pi \epsilon_0} \frac{q}{(r+a)^2}$ (along $\vec{PA}$)

Electric field  Intensity at a Point P due to B(+q) is

$\large E_{PB} = \frac{1}{4\pi \epsilon_0} \frac{q}{BP^2}$

$\large E_{PB} = \frac{1}{4\pi \epsilon_0} \frac{q}{(r-a)^2}$ (along $\vec{BP} \; produced $)

As both the electric field intensity are in opposite direction .

Net Electric field Intensity at P is

$\large E = E_{PB} – E_{PA}$

$\large E = \frac{q}{4\pi \epsilon_0} [\frac{1}{(r-a)^2} – \frac{1}{(r+a)^2}] $

$\large E = \frac{q}{4\pi \epsilon_0} [\frac{(r+a)^2 – (r-a)^2}{(r^2-a^2)^2} ] $

$\large E = \frac{q}{4\pi \epsilon_0} [\frac{4 r a}{(r^2-a^2)^2} ] $

$\large E = \frac{1}{4\pi \epsilon_0} [\frac{(q \times 2a) 2 r}{(r^2-a^2)^2} ] $

$\large E = \frac{1}{4\pi \epsilon_0} [\frac{ 2 p r}{(r^2-a^2)^2} ] $ (Where q × 2a = p = dipole moment)

If dipole is very short then a<< r

$\large E = \frac{1}{4\pi \epsilon_0} [\frac{2 p}{r^3} ] $

Note : The angle between Electric field intensity & dipole moment is 0°

At a Point on equitorial  Line :

We have to find Electric field intensity at Point P .

Electric field  Intensity at a Point P due to A(-q) is

$\large E_{PA} = \frac{1}{4\pi \epsilon_0} \frac{q}{AP^2}$

$\large E_{PA} = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2 + a^2}$ (along $\vec{PA}$)

Electric field  Intensity at a Point P due to B(+q) is

$\large E_{PB} = \frac{1}{4\pi \epsilon_0} \frac{q}{BP^2}$

$\large E_{PB} = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2 + a^2}$ (along $\vec{BP} \; produced $)

Here , $\large E_{PA} = E_{PB}$

By resolving these field intensity , $\large E_{PA} sin\theta \; and \;  E_{PB} sin\theta $  will be cancel out .

Net Electric field Intensity at P is

$\large E = E_{PA} cos\theta + E_{PB} cos\theta $

$\large E = 2 E_{PA} cos\theta  $

$\large E = 2 \frac{1}{4\pi \epsilon_0}\frac{q}{(r^2 + a^2)}. \frac{a}{\sqrt{r^2 + a^2}} $

$\large E = 2 \frac{1}{4\pi \epsilon_0}\frac{q \times 2a}{(r^2 + a^2)^{3/2}} $

$\large E = 2 \frac{1}{4\pi \epsilon_0}\frac{p}{(r^2 + a^2)^{3/2}} $

If dipole is short then , a<< r

$\large E = 2 \frac{1}{4\pi \epsilon_0}\frac{p}{r^3} $

Note : The angle between electric field Intensity & dipole moment is 180°

 Electric field due to the dipole at point P (r,θ):

Electric field due to the dipole at point P (r,θ)is given by

$ \displaystyle \vec{E} = \frac{2p cos\theta}{4 \pi \epsilon_0 r^3}\, \hat{r} + \frac{p sin\theta}{4 \pi \epsilon_0 r^3} \, \hat{\theta} $    ; where r >> a

The direction of unit vector r^ is same as of r whereas the unit vector θ^ is perpendicular to r in the direction of increasing θ .

Along the dipole axis, θ = 0° . Thus

$ \displaystyle \vec{E} = \frac{2p cos\theta}{4 \pi \epsilon_0 r^3}\, \hat{r} $

Along the perpendicular bisector, θ = 90 °  . Thus

$ \displaystyle \vec{E} = \frac{p sin\theta}{4 \pi \epsilon_0 r^3} \, \hat{\theta} $

(iii) Potential due to a dipole at any point P(r , θ) is given by

$ \displaystyle   V = \frac{p cos\theta}{4 \pi \epsilon_0  r^2} $

Torque acting on a dipole placed in a uniform electric field:

When a dipole is kept in an electric field , in general a torque and a force act on it.

But, if the field is uniform the net force becomes zero and the torque is given by

$\large \tau = qE \times Perpendicular \; distance $

$\large \tau = qE \times 2a sin\theta $

$\large \tau = (q \times 2a )E  sin\theta $

$\large \tau = pE  sin\theta $ ; where p = electric dipole moment

$ \large \vec{\tau} = \vec{p} \times \vec{E} $

 Potential energy of dipole in a uniform electric field :

Small amount of work done in rotating a dipole through small angle dθ is

$\large dw = \tau d\theta$

Work done in rotating a dipole from θ1 to θ2 is

$\large W = \int_{\theta_1}^{\theta_2} \tau d\theta$

$\large W = \int_{\theta_1}^{\theta_2} p E sin\theta d\theta $

$\large W = p E \int_{\theta_1}^{\theta_2} sin\theta d\theta $

$\large W = p E [-cos\theta]_{\theta_1}^{\theta_2}$

$\large W = -p E [cos\theta_2 – cos\theta_1]  $

If Potential Energy is arbitrarily taken zero when dipole is at 90° , then potential energy in rotating the dipole and inclining at an angle θ is

Potential Energy ,  $\large U = -p E cos\theta  $

$ \displaystyle U = -\vec{p}.\vec{E} $

Note :

case(i) If θ = 0° , then U = -p E (Stable equilibrium)

case(ii) If θ = 90° , then U = 0

case(i) If θ = 180° , then U = p E (Unstable equilibrium)

Example: Two tiny spheres, each of of mass M , and charges +q and -q respectively , are connected by a massless rod of length , L . They are placed in a uniform electric field at an angle θ with the E (θ ≈ 0o). Calculate the minimum time in which the dipole axis becomes parallel to the field line.

Solution:

τ = p E sin θ, (as for small angle ,sin θ → θ)

⇒ τ = − p E θ (If we assume angular displacement to be anti-clockwise, torque is clockwise)

As torque is proportional to ‘θ’ and oppositely directed, the motion will be an angular S.H.M.

⇒ τ = I α

I α = − p E θ

$ \displaystyle \alpha = -\frac{p E}{I} \theta $

As , α = – ω2θ

$ \displaystyle \omega = \sqrt{\frac{p E}{I} } $

Here, dipole moment , p = q × L

Moment of inertia , $ \displaystyle I = M (\frac{L}{2})^2 + M (\frac{L}{2})^2 $

Time period $ \displaystyle T = \frac{2\pi}{\omega} $

$ \displaystyle T = 2\pi \sqrt{\frac{I}{p E}} $

The minimum time to required align itself is T/4 sec

Exercise : A dipole of dipole moment p is kept along an electric field E such that p and E are in the same direction. Find the work done in rotating the dipole by an angle π .

Also Read :

Charging of the body & Properties of Charge
Coulom’s Law in Electrostatics
Principle of superpostion
Electric field
Electric Potential Energy
Electric Lines of Force & its Properties
Electric Potential
Equipotential Surface , Relation b/w Electric field & Potential
Electric Flux , Gauss’ Law & Applications of Gauss’s Law
Charge appearing on the Surface of Plates
Mechanical Force & Electric Pressure on a Charged Surface

Next Page → 

← Back Page