Physics Topics such as mechanics, thermodynamics, and electromagnetism are fundamental to many other scientific fields.
What is Electric Dipole and its Formula?
Definition: An electric dipole is a combination of two equal but opposite point charges separated by a small distance.
Example: The molecules of water (H2O), ammonia (NH3), hydrochloric acid (HCl), carbon dioxide (CO2) etc. are electric dipoles. A water molecule and a hydrochloric acid molecule are shown in Fig.
The total charge of the electric dipole is zero (as, +q – q = 0), but the field of the electric dipole is not zero.
Dipole moment: The moment of an electric dipole is the product of the magnitude of either charge (q) and the distance (2l) between them.
According to Fig., dipole moment, p = q ᐧ 2l ………. (1)
Electric dipole moment is a vector quantity. It is directed along the axis of the dipole (the line joining the two charges) from the negative to the positive charge.
The vector form of dipole moment is \(\vec{p}\) = 2q\(\vec{l}\)
Unit and dimension: In CGS system the unit of dipole moment is stateC ᐧ cm and in SI it is C ᐧ m. Its dimension is [p] = LTI.
Field Intensity at a Point on the Axis of a Dipole
Let the two charges -q and +q separated by a distance 21 form an electric dipole AB [Fig.]. Clearly AB = 2l and dipole moment, p = 2lq. Electric field intensity at a point P situated on the axis of the dipole is to be calculated. This position of P with respect to the dipole is called end-on position or axial position.
Let the distance of the point P from the centre O of the diople be r.
Suppose, the dipole is placed in a medium of dielectric constant k. The permittivity of air medium or vacuum is ε0.
Now, field intensity at P due to the charge +q,
E1 = \(\frac{q}{4 \pi \kappa \epsilon_0(r-l)^2}\) ; along \(\overrightarrow{O P}\)
Again, field intensity at P due to the charge – q,
E2 = \(\frac{q}{4 \pi \kappa \epsilon_0(r+l)^2}\) ; along \(\overrightarrow{P O}\)
As E1 and E2 acting in opposite directions and E1 > E2, there-fore, resultant field intensity at P,
E = E1 – E2
Obviously, the direction of E is along \(\overrightarrow{O P}\).
For air or vacuum, k = 1
∴ E = \(\frac{2 p r}{4 \pi \epsilon_0\left(r^2-l^2\right)^2}\) ….. (2)
The vector form of equation (1) is
\(\vec{E}\) = \(\frac{2 \overrightarrow{p r}}{4 \pi \kappa \epsilon_0\left(r^2-l^2\right)^2}\)
The corresponding CGS expression of equation (2) is
E = \(\frac{2 p r}{\left(r^2-l^2\right)^2}\)
Special case: If the point P is far away from the centre O of the dipole i.e., if r»l, then neglecting l2 in comparison with r2, we get from equations (1) and (2) respectively,
E = \(\frac{1}{4 \pi k \epsilon_0} \cdot \frac{2 p}{r^3}\) … (3)
and E = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{2 p}{r^3}\) … (4)
For this end-on position, \(\vec{p}\) and \(\vec{E}\) are along the same direc-tion.
Field Intensity at a Point on the Perpendicular Bisector of a Dipole
Any line passing through the centre of the dipole and perpendic-ular to its axis is called its perpendicular bisector.
Let the two charges -q and +q separated by a distance 21 form an electric dipole AB [Fig.]. Clearly AB = 2l and dipole moment, p = 2lq. Electric field intensity at a point P situated on the perpendicular bisector of the dipole is to be calculated. This position of P with respect to the dipole is called broadside-on or equatorial position.
Let the distance of the point P from the centre of the dipole be r.
Suppose, the dipole is placed in a medium of dielectric constant k.
The permittivity of air medium or vacuum is ε0.
Now, field intensity at P due to the charge +q,
E1 = \(\frac{1}{4 \pi \kappa \epsilon_0} \cdot \frac{q}{(B P)^2}\) ; along \(\overrightarrow{B P}\)
Again, field intensity at P due to the charge -q,
E2 = \(\frac{1}{4 \pi \kappa \epsilon_0} \cdot \frac{q}{(A P)^2}\) ; along \(\overrightarrow{P A}\)
Since BP = AP
∴ E1 = E2 and ∠PAB = ∠PBA = θ (say)
If PR be parallel to BA, then
∠QPR = ∠PBA = d and ∠RPA = ∠PAB = 6
Now, the field intensity E1 is resolved into two perpendicular components along PR and PT (extended OP). The components are respectively E1 cos θ and E1 sin θ.
Similarly, the field intensity E2 is resolved into two perpendicular components along PR and PO. The components are respec-tively E2 cosθ and E2sinθ.
The sine components along PT and PO being equal and oppo-site balance each other. So the resultant field intensity at P is
Special cose: If the point P is far away from the centre O of the dipole i.e., if r\(\gg\)1, then neglecting l2 in comparison with r2, we get from equations (1) and (2) respectively,
E = \(\frac{1}{4 \pi k \epsilon_0} \cdot \frac{p}{r^3}\) …. (3)
and E = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{r^3}\) …. (4)
For this broadside-on position, and are oppositely directed.
It is to be noted that, when point P is far away from the centre of the dipole, the electric field intensity at a point on axial position is double the electric field intensity at a point on the equatorial position.
Field Intensity at any Point due to an Electric Dipole
Suppose, the point P is situated at a distance r from the centre O of the dipole and the line OP makes an angle θ with the axis of the dipole. The dipole moment \(\vec{p}\) acts along \(\overrightarrow{A B}\) and the length of the dipole is negligible in comparison to r [Fig.].
\(\vec{p}\) is resolved into two perpendicular components:
i) p cosθ; along \(\overrightarrow{O P}\)
ii) p sinθ; along perpendicular to \(\overrightarrow{O P}\)
Hence, the point P is situated on the axial line of a dipole of moment pcosθ, and on the equatorial line of a dipole of moment psinθ.
Let \(\vec{E}_1\) and \(\vec{E}_2\) be the intensities at P due to the components pcosθ and psinθ respectively, and the resultant intensity at P be \(\vec{E}\).
So, \(\vec{E}\) = \(\vec{E}_1\) + \(\vec{E}_2\)
∴ E2 = \(E_1^2\) + \(E_2^2\) or, E = \(\sqrt{E_1^2+E_2^2}\)
Suppose, the dipole is placed in a medium of dielectric constant K . The permittivity of air medium or vacuum is ε0.
From equation (3) in section 2.4.1 and equation (3) in section 2.4.2, we get respectively, p
Special cases:
i) End-on position: If the point P is on the end-on position, i.e., on the axis of the dipole, # = 0° or cos# = 1.
Therefore, from equation (1) we get,
E = \(\frac{1}{4 \pi \kappa \epsilon_0} \cdot \frac{p}{r^3} \sqrt{3+1}\) = \(\frac{1}{4 \pi \kappa \epsilon_0} \cdot \frac{2 p}{r^3}\)
This is equation (3) in section 2.4.1.
ii) Broadside-on position or equational position: If the point P is on the broadside-on position, i.e., on the perpendicular bisector of the dipole, θ = 90° or, cosθ = 0.
Therefore, from equation (1) we get,
E = \(\frac{1}{4 \pi \kappa \epsilon_0} \cdot \frac{p}{r^3} \sqrt{0+1}\) = \(\frac{1}{4 \pi \kappa \epsilon_0} \cdot \frac{p}{r^3}\)
This is equation (3) in section 2.4.2.
Torque Acting on an Electric Dipole in a Uniform Electric Field
Consider an electric dipole consisting of charges -q at A and + q at B separated by a distance 2l [Fig.], So dipole moment, p = q ᐧ 2l. It is a vector quantity and its direction is from negative to positive charge along the axis of the dipole.
Suppose, the dipole makes an angle θ with a uniform electric field of strength E. The force on charge +q is qE, along the direction of \(\vec{E}\) and force on charge -q is qE, opposite to the direction of \(\vec{E}\). Obviously, these forces are equal in magnitude but opposite in direction. Therefore, net force on the dipole, \(\vec{F}_{\text {net }}\) = \(q \vec{E}\) – \(q \vec{E}\) = 0 i.e., net force on electric dipole in uni-form electric field is zero.
Since these two forces are equal in magnitude, opposite in direc-tion and act at different points, they constitute a couple. If the magnitude of the torque is r, then
\(\tau\) = either force × perpendicular distance between the lines of action of the two forces
= qE × AC
= qE × 2lsinθ = q × 2l × Esinθ or, \(\tau\) = pEsinθ …… (1)
The torque acting on the dipole tends to bring it along the direc-tion of the electric field.
Special cases:
i) When θ = 90°, the magnitude of the torque becomes max-imum, i.e., \(\tau\) = pE .In this case, the electric dipole is in sta-ble equilibrium.
Now, if E = 1, \(\tau\) = p = dipole moment
It provides an alternative definition of electric dipole moment:
The moment of an electric dipole is defined as the torque acting on it when it is placed at right angles to a uniform electric field of unit strength.
ii) When θ = 0° or 180°, the magnitude of the torque is min-imum, i.e., \(\tau\) = 0 . In this case, the electric dipole is in an unstable equilibrium.
Torque in vector form: Torque \(\vec{\tau}\) acting on a dipole of moment \(\vec{p}\) when placed in a uniform electric field \(\vec{E}\) is given in vector form as,
\(\vec{\tau}\) = \(\vec{p} \times \vec{E}\) …… (2)
According to the rule of cross product, \(\vec{\tau}\) is perpendicular to the plane containing \(\vec{p}\) and \(\vec{E}\) [Fig.].
Numerical Examples
Example 1.
How many field lines would emanate from a 1 C posi-tive charge placed in vacuum?
Solution:
Number of lines emanated
= \(\frac{1}{\epsilon_0}\) = \(\frac{1}{8.854 \times 10^{-12}}\) = 1.129 × 1011
Example 2.
A straight rod of length l placed in vacuum is charged uniformly with an amount q of charge. Calculate the electric field intensity in SI at a point on the axis of the rod at a distance x from its nearer end. Write the result in CGS system also.
Solution:
The length of the rod AB is Z. Let P be a point at a distance x from the end B [Fig.]. Since the rod is uniformly charged with q, the linear charge density of the rod is λ = \(\frac{q}{l}\).
Consider a small length dr of the rod at a distance r from P. Charge of this portion of length dr,
dq = λdr = \(\frac{q}{l}\)dr
Electric field intensity at P due to this charge dq,
dE = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{\frac{q}{l}}{r^2} d r\) = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{l} \frac{d r}{r^2}\)
Therefore, field intensity at P due to the whole charged rod,
In CGS system, replacing ε0 by \(\frac{1}{4 \pi}\) the electric field intensity becomes
E = \(\frac{q}{x(l+x)}\) ;along \(\overrightarrow{B P}\)
Example 3.
Two point charges are placed on the y -axis at y = +a and y = -a in vacuum. The magnitude of each charge is q. Determine the electric field Intensity at P(x, 0) on the x -axis.
Solution:
Suppose, electric field intensity at the point P due to the charges at A and B are E1 and E2, respectively [Fig.].
Both E1 and E2 are resolved into two perpendicular compo-nents. It is seen that the sine components being equal and oppo-site, balance each other. But the cosine components are added up as they act in the same direction.
∴ Field strength at P
= 2Ecosθ = 2 × \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{\left(a^2+x^2\right)} \cdot \frac{x}{\sqrt{a^2+x^2}}\)
= \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{2 q_x}{\left(a^2+x^2\right)^{3 / 2}}\)
Example 4.
An electric dipole placed in vacuum is formed by two equal but opposite charges each of magnitude 1µC separated by a distance of 2 cm. Calculate the electric field intensities in the following cases :
(i) at a point on the axis of the dipole situated at a distance 60 cm away from its centre.
(ii) at a point on the perpendicular bisector of the dipole situated at a distance 60 cm away from its centre.
Solution:
Moment of the electric dipole,
p = q ᐧ 2l = 1 × 10-6 × 2 × 10-2 = 2 × 10-8 C ᐧ m ]
i) Electric field intensity on the axis of the dipole,
E1 = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{2 p}{r^3}\) = \(\frac{9 \times 10^9 \times 2 \times 2 \times 10^{-8}}{(0.6)^3}\)
= 1666.6 N ᐧ C-1 ; along the axis
ii) Electric field intensity on the perpendicular bisector of the dipole,
E2 = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{r^3}\) = \(\frac{1}{2} E_1\)
= 833.3 C-1; parallel to axis of the dipole
Example 5.
Three charges q, -2q and q are placed at three verti-ces of an equilateral triangle. Determine the equiva-lent dipole moment of the system.
Solution:
The given system is a combination at two dipoles at an angle 60° with each other. [Fig.]
Let the dipole moment of one dipole is p1 and that of another dipole is p2, the length of each side of the equilateral triangle is a. Equivalent dipole moment,
p = P1cos30°+ p2cos30°
[∵ sine components cancel each other]
= 2qa cos 30° [∵ p1 = p2 = qa]
= 2qa × \(\frac{\sqrt{3}}{2}\) = \(\sqrt{3} q a\)
Example 6.
Electric field intensities at two axial points of an elec-tric dipole at distances 5 cm and 10 cm from its centre are 2.5 × 104 N ᐧ C-1 and 2 × 103 N ᐧ C-1 respectively. The dipole is placed in air. Determine the length of the dipole.
Solution:
Suppose, the length of the electric dipole = 2l
Example 7.
An electric dipole is formed by two equal but opposite charges, each of magnitude 1µC, separated by a distance of 5 cm. What is the magnitude of the torque required to place the dipole at right angles to an electric field of intensity 3 × 105 N ᐧ C-1 ?
Solution:
Here,
p = q × 2l = l × 10-6 × 5 × 10-2 = 5 × 10-8 C ᐧ m E = 3 × 105 N ᐧ C-1 and θ = 90°
∴ The required torque,
r = pEsinθ = 5 × 10-8 × 3 × 105 × sin90°
= 0.015 N ᐧ m
Example 8.
An electric dipole is formed by placing charges ±20 × 10-6 C at a distance 2 mm. Calculate the electric field at a point on the perpendicular bisector of the axis of the dipole situated at a distance 10 cm from the midpoint of the dipole. [HS ’11]
Solution:
Electric Field, E = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{\left(r^2+l^2\right)^{3 / 2}}\)
[here, l = 1 × 10-3m, p = q × 2l
= 20 × 10-6 × 2 × 10-3 = 40 × 10-9 cm]
∴ E = \(\frac{9 \times 10^9 \times 40 \times 10^{-9}}{\left[(10 \times 10)^{-2}+\left(1 \times 10^{-3}\right)^2\right]^{3 / 2}}\) = \(\frac{360}{\left(10^{-2}+10^{-6}\right)^{3 / 2}}\)
= 36 × 104 N ᐧ C-1