Informative line

Dipole

Practice to calculate torque on a dipole in a uniform field and Potential Energy due to Dipole-Dipole Interaction. Learn definition and formula of dipole moment with examples.

Dipole

Definition-

  • Two equal and opposite point charges separated by a very small distance is called dipole.

Dipole Moment

  • Dipole moment is the multiplication of one charge and distance between them.

\(\vec p=q(\vec d)\)

S.I. unit of dipole moment

Unit:- C-m

Direction of dipole moment

  • Direction of dipole moment is always taken from negative charge to positive charge.
  • Dipole moment is a vector quantity.

Illustration Questions

Which one of the following dipoles does not have  magnitude of dipole moment p=24 nC-m? 

A

B

C

D

×

For option (A),

Dipole moment p = q(d)

\(\Rightarrow 2\times10^{-6}\times12\times10^{-3}=24\,\text{nC-m}\)

 

For option (B),

Dipole moment p = q(d) 

\(\Rightarrow3\times10^{-6}\times8\times10^{-3}=24\;\text{nC-m}\)

 

For option (C),

Dipole moment p = q(d) 

\(\Rightarrow12\times10^{-6}\times2\times10^{-3}=24\;\text{nC-m}\)

For option (D),

Dipole moment p = q(d)

\(\Rightarrow1\times10^{-6}\times3\times10^{-3}=3\;\text{nC-m}\)

Which one of the following dipoles does not have  magnitude of dipole moment p=24 nC-m? 

A image
B image
C image
D image

Option D is Correct

Electric Field on the Axis and Equator of Dipole

  • Consider a dipole shown in figure.

Calculation of Electric Field on Axial Line

  • Consider a point P on the axis of dipole at a distance r from the mid-point of the dipole.

\(\vec E_{-q}=\dfrac {1}{4\pi\epsilon_0}\dfrac {q}{(r+\ell)^2}\)    (in negative X-axis)

\(\vec E_{+q}=\dfrac {1}{4\pi\epsilon_0}\dfrac {q}{(r-\ell)^2}\)    ( in positive X-axis)

Since, \((r+\ell)^2>(r-\ell)^2\)

so, \(|\vec E_{-q}|<|\vec E_{+q}|\) 

  • So, at point P net field will be in positive X-axis.

\(\vec E_{net}=E_{+q}-E_{-q}\)       (towards positive X-axis)

\(\vec E_{net}= \dfrac {1}{4\pi\epsilon_0} \dfrac {q}{(r-\ell)^2}- \dfrac {1}{4\pi\epsilon_0} \dfrac {q}{(r+\ell)^2}\)     (towards positive X-axis)

\(\vec E_{net}= \dfrac {q}{4\pi\epsilon_0} \left [ \dfrac {1}{(r-\ell)^2}- \dfrac {1}{(r+\ell)^2} \right]\)   (towards positive X-axis)

\(\vec E_{net}= \dfrac {q}{4\pi\epsilon_0} \left [ \dfrac {4r\ell}{(r^2-\ell^2)^2} \right]\) (towards positive X-axis)

  • Dipole moment

\(\vec p=q(2 \ell)\) (towards positive X-axis)

\(\vec E_{net}=\dfrac {2\vec pr}{4\pi\epsilon_0(r^2-\ell^2)^2}\)

  • If \(\ell <<r\)

\(\vec E_{net}=\dfrac {2\vec pr}{4\pi\epsilon_0(r^2)^2}\)

\(\vec E_{net}=\dfrac {2\vec p}{4\pi\epsilon_0\;r^3}\)

Calculation of Electric Field on Equatorial Line of Dipole

  • By superposition principle, total electric field at R is the vector sum of electric field of -q and +q at R.

\(E_{net}=2\;E cos\;\theta\) (in opposite direction to \(\vec p\))

\(E_{net}=2. \dfrac {1}{4\pi\epsilon_0} \dfrac {q}{(r^2+\ell^2)} cos\theta \)

 

 

\(cos\theta = \dfrac {\ell}{\sqrt{r^2+\ell^2}}\)

\(E_{net}=\dfrac {1}{4\pi\epsilon_0} \dfrac {2q\ell}{(r^2+\ell^2)^{3/2}} \) (opposite direction to \(\vec d\))

Since \(\ell <<r\)

        \(\therefore r^2+\ell^2\simeq r^2\)

\(\vec E_{net}=\dfrac {1}{4\pi\epsilon_0}\times\dfrac {(-\vec p)}{r^3}\)

\(\vec E_{net}=\dfrac {-\vec p}{4\pi\epsilon_0\;r^3}\)

Illustration Questions

Choose the incorrect expression for electric field at a point on the axial  and equatorial line if, charges are +q and –q and distance between them is d.

A \((\vec E_{P\,})_{Axial}=-\dfrac {1}{4\pi\epsilon_0} \dfrac{q\vec d}{r^3}\),  \((\vec E_{P\,})_{Equatorial}=\dfrac {+\vec p}{4\pi\epsilon_0r^3}\) where, \(\vec p=q\,\vec d\) 

B \((\vec E_{P\,})_{Axial}=+\dfrac {1}{4\pi\epsilon_0}\times \dfrac{2q\vec d}{r^3}\),  \((\vec E_{P\,})_{Equatorial}=\dfrac {-q\vec d}{4\pi\epsilon_0r^3}\) where, \(\vec p=q\,\vec d\) 

C \((\vec E_{P\,})_{Axial}=\dfrac {1}{4\pi\epsilon_0}\times \dfrac{2\vec p}{r^3}\),  \((\vec E_{P\,})_{Equatorial}=\dfrac {-\vec p}{4\pi\epsilon_0r^3}\) where, \(\vec p=q\,\vec d\) 

D \((\vec E_{P\,})_{Axial}=\dfrac {1}{2\pi\epsilon_0}\times \dfrac{q\vec d}{r^3}\),  \((\vec E_{P\,})_{Equatorial}=\dfrac {-\vec p}{4\pi\epsilon_0r^3}\) where, \(\vec p=q\,\vec d\) 

×

Electric field at a point on axial line \((\vec E_P)_{Axial}=\dfrac {1}{4\pi\epsilon_0}\dfrac {2\vec p}{r^3}\)

Electric field at a point on equatorial line \((\vec E_P)_{Equatorial}=\dfrac {-1}{4\pi\epsilon_0}\dfrac {\vec p}{r^3}\)(where \(\vec p = q\vec d\) )

Hence, option (A) is incorrect.

Choose the incorrect expression for electric field at a point on the axial  and equatorial line if, charges are +q and –q and distance between them is d.

image
A

\((\vec E_{P\,})_{Axial}=-\dfrac {1}{4\pi\epsilon_0} \dfrac{q\vec d}{r^3}\)

\((\vec E_{P\,})_{Equatorial}=\dfrac {+\vec p}{4\pi\epsilon_0r^3}\)

where, \(\vec p=q\,\vec d\) 

.

B

\((\vec E_{P\,})_{Axial}=+\dfrac {1}{4\pi\epsilon_0}\times \dfrac{2q\vec d}{r^3}\),

 \((\vec E_{P\,})_{Equatorial}=\dfrac {-q\vec d}{4\pi\epsilon_0r^3}\)

where, \(\vec p=q\,\vec d\) 

C

\((\vec E_{P\,})_{Axial}=\dfrac {1}{4\pi\epsilon_0}\times \dfrac{2\vec p}{r^3}\),

 \((\vec E_{P\,})_{Equatorial}=\dfrac {-\vec p}{4\pi\epsilon_0r^3}\)

where, \(\vec p=q\,\vec d\) 

D

\((\vec E_{P\,})_{Axial}=\dfrac {1}{2\pi\epsilon_0}\times \dfrac{q\vec d}{r^3}\),

 \((\vec E_{P\,})_{Equatorial}=\dfrac {-\vec p}{4\pi\epsilon_0r^3}\)

where, \(\vec p=q\,\vec d\) 

Option A is Correct

Electric Potential on the Axis of a Dipole

  • Consider a dipole as shown in figure.

 

  • Total electric potential at point P is the scalar sum of potential due to –q and +q charge.

\(V_p=V_{-q}+V_{+q}\)

\(V_p= \dfrac {1}{4\pi\epsilon_0} \dfrac {-q}{(r+\ell)} + \dfrac {1}{4\pi\epsilon_0} \dfrac {q}{(r-\ell)}\)

\(V_p= \dfrac {1}{4\pi\epsilon_0}q \left [ \dfrac {1}{(r-\ell)} - \dfrac {1}{(r+\ell)} \right]\)

\(V_p= \dfrac {1}{4\pi\epsilon_0}q \left [ \dfrac {(r+\ell)-{(r-\ell)}} {{(r+\ell)}(r-\ell)} \right]\)

\(V_p= \dfrac {1}{4\pi\epsilon_0}q \left [ \dfrac {2\ell} {r^2-\ell^2} \right]\)

Since, \(\ell <<r\)  (\(r^2-\ell^2\simeq r^2\))

\(V_P=\dfrac {1}{4\pi\epsilon_0} \dfrac {|\vec p|}{r^2}\)

Note- A dipole having zero potential at equatorial line.

Illustration Questions

Calculate potential at point P due to dipole as shown in figure. Given, d = 1mm, Q=2\(\mu\)C and r = 2m.

A 65.4 Volt

B 60 Volt

C 15 Volt

D 4.5 Volt

×

Potential at point P,

\(V_P=\dfrac {1}{4\pi\epsilon_0} \dfrac {|\vec p|}{r^2}\)

where,

p = dipole moment

r = separation from mid point of dipole to point P

image

\(\dfrac {1}{4\pi\epsilon_0} = 9\times10^9\;\text{Nm}^2/\text C^2\)

\(|\vec p|=q\times d = (2\times 10^{-6})\times (1\times 10^{-3}) \)

\(= 2\times 10^{-9}\)

\(r = 2\,m\)

 

image

\(V_P=\dfrac {9\times 10^9\times2\times10^{-9}}{(2)^2}\)

\(V_P=4.5\;\text{ Volt}\)

image

Calculate potential at point P due to dipole as shown in figure. Given, d = 1mm, Q=2\(\mu\)C and r = 2m.

image
A

65.4 Volt

.

B

60 Volt

C

15 Volt

D

4.5 Volt

Option D is Correct

Torque Experienced by a Dipole in a Uniform Electric Field

  • Consider a dipole in a uniform electric field (\(\vec E\)), as shown in figure.

  • The dipole is placed, making an angle \(\theta\) with the field vector.
  • By analyzing forces on both the point charges, it is clear that the net force on the dipole is zero.

\(\vec F_{Net}=\vec F_{+q}+\vec F _{-q}\)

\(\vec F_{Net}=q\vec E+(-q\vec E)\)

\(\vec F_{Net}=0\)

 

  • But due to the force couple, a torque is created.

\(\tau\)= Couple force × Couple arm

\(\tau= qE (d sin\,\theta)\)

\(\tau=(qd)\;(E)sin\;\theta\)

\(\vec \tau=q\vec d\times \vec E\)

\(\vec \tau=\vec p\times \vec E\)

Maxima and Minima Condition of Torque

Maxima

  • Torque will be maximum when dipole is perpendicularly placed in an electric field.

\(\theta = 90°\)

\(\tau_{maximum}=q\;Ed=pE\)

Minima

  • Torque will be minimum when dipole is placed in the direction of electric field or in the direction opposite to electric field.
  • In figure, \(\tau_{minimum}=0\) . 

Note :  If \(\vec E\) is non-uniform, expression of torque experienced by dipole changes.

Illustration Questions

A uniform electric field of intensity \(\vec E=10^8\) Volt/m is directed in X-Y plane along X-axis. Calculate the torque experienced by a dipole as shown in figure. (Given d = 2 mm, \(\theta\)=30°, Q = 2 \(\mu\)C)

A 390 N-m

B 1200 N-m

C 3.9 N-m

D 0.2 N-m

×

Net torque on dipole,

\(\tau=(q\;d)E\;sin\theta\)

\(\tau =(2\times10^{-6})\;(2\times10^{-3})\;10^8\left(\dfrac {1}{2}\right)\)

\(\tau =2\times10^{-1}\)

\(\tau =0.2\,\text{N-m}\)

image

A uniform electric field of intensity \(\vec E=10^8\) Volt/m is directed in X-Y plane along X-axis. Calculate the torque experienced by a dipole as shown in figure. (Given d = 2 mm, \(\theta\)=30°, Q = 2 \(\mu\)C)

image
A

390 N-m

.

B

1200 N-m

C

3.9 N-m

D

0.2 N-m

Option D is Correct

Electric Field at a Point

  • To find electric field at a point (P) which lies neither on the axis nor on the equatorial line, electric field due to -q and +q at P can be added vectorially to get net field at P.

Note: This method is very complicated.

  • Other method is to take components of dipole moments of the dipole such that point P lies on the axis of one component of dipole and on the equator of other component of dipole.

  • Electric field due to component of dipole moment is in direction of dipole moment component.
  • Now at point P,

Reason\(\rightarrow\)

(1) Axial line

(2) Equator Line

  • Net electric field at P,

  • Net electric field at point P,

\(\vec E = \dfrac {p}{4\pi\epsilon_0r^3}(2cos^2\theta-sin^2\theta)\hat i+\dfrac {3p\,sin\theta \,cos\theta}{4\pi\epsilon_0r^3}\hat j\)

 

 

Illustration Questions

Calculate electric field at point A, at a distance r = 5 m and making an angle \(\theta\)= 30°, from the mid point of dipole having charge q = 2 \(\mu\) C and distance between charges d = 2 mm.

A \((39\hat i +37\hat j)\,V/m\)

B \((10\hat i +15\hat j)\,V/m\)

C \((0.35\hat i +0.325\hat j)\,V/m\)

D \((18\hat i +12\hat j)\,V/m\)

×

Dipole moment 

\(\vec p=2\times10^{-6}\times2\times 10^{-3}(\hat i)\,\text{C-m}\)

\(\vec p=4\times10^{-9}(\hat i)\,\text{C-m}\)

image

Taking components of this dipole moment such that point A lies on the axis of one component and the equator of other component.

image

Net field at point A ,

image

At point A along equatorial line ,

\( E\;psin\,\theta=\dfrac {1\times p}{4\pi\epsilon_0r^3}sin30°\)

\( E\;psin\,\theta=\dfrac {4\times 10^{-9}\times 9\times10^{9}}{125}\times\dfrac {1}{2}\)

\(=\dfrac {18}{125}=0.15\)

At point A along axial line ,

\(\vec E\;pcos\,\theta=\dfrac {2\times p}{4\pi\epsilon_0r^3}cos30°\)

\( Ep\,cos\,\theta=\dfrac {2\times4\times 10^{-9}} {125}\times\dfrac {\sqrt3}{2}\times9\times10^9\)

\(=\dfrac {36\sqrt3}{125}=0.5\)

Net field at point A,

image

Net field at A,

image

Net field at A,

image

\(\vec E=(0.35\hat i+0.325 \hat j)\,V/m\)

Calculate electric field at point A, at a distance r = 5 m and making an angle \(\theta\)= 30°, from the mid point of dipole having charge q = 2 \(\mu\) C and distance between charges d = 2 mm.

image
A

\((39\hat i +37\hat j)\,V/m\)

.

B

\((10\hat i +15\hat j)\,V/m\)

C

\((0.35\hat i +0.325\hat j)\,V/m\)

D

\((18\hat i +12\hat j)\,V/m\)

Option C is Correct

Potential at a Point

  • Total potential at P is the scalar sum of potential due to –q and +q.

\(V_P=(V_P)_{-q} +(V_P)_{+q}\)

Note : Above method is very complicated.

  • Other way is to take components of dipole moment of the dipole such that point P lies on the axis of one component and on the equatorial line of other component.

  • For \(p\;cos\theta\), point P lies on the axial. So, potential at point P,

\((V_P)_{p\,cos\theta}=\dfrac {1}{4\pi\epsilon_0}\times\dfrac {p\, cos\,\theta}{r^2}\)

  • For \(p\,\sin\theta\), point P lies on the equatorial line so, potential at P due to p sin \(\theta\)

\((V_P)_{p\,sin\theta}=0\)

  • Total potential at point P,

\(V=\dfrac {1}{4\pi\epsilon_0}\times\dfrac {p\,cos\,\theta}{r^2}\)

Illustration Questions

Calculate electric potential at  point A, as shown in figure. Given \(\theta = \)60°, r = 5 m, d = 2 mm, q = 5 \(\mu\)C, –q =–5 \(\mu\)C.

A –4.6 Volt

B 1.8 Volt

C 3.9 Volt

D 9.2 Volt

×

Dipole moment (\((\vec p)=q\times d\))

\(\vec p=5\times10^{-6}\times2\times 10^{-3}\hat i\)

\(\vec p=10\times10^{-9}\hat i\) C-m

image

Taking components of dipole moment 

image image

Dipole moment along axial line = p cos 60°

 \(=10 \times 10^{-9}\times \dfrac {1}{2}=\dfrac {10^{-8}}{2}\)

Dipole moment along equatorial line = p sin 60°

 \(=10 \times 10^{-9}\times \dfrac {\sqrt3}{2}=5\sqrt 3\times 10^{-9}\)

image image

\((V_P)_{cos\;60°_A}=\dfrac {1}{4\pi\epsilon_0}\times\dfrac {p\,cos\,60°}{r^2}\)

\(=\dfrac {9\times10^9\times10^{-8}}{(5)^2}\times\dfrac {1}{2}\)

\(= 1.8 \,Volt\)

\((V_P)_{sin\;60°_A}=0\)

image

Total potential = 1.8 + 0

= 1.8 Volt

image

Calculate electric potential at  point A, as shown in figure. Given \(\theta = \)60°, r = 5 m, d = 2 mm, q = 5 \(\mu\)C, –q =–5 \(\mu\)C.

image
A

–4.6 Volt

.

B

1.8 Volt

C

3.9 Volt

D

9.2 Volt

Option B is Correct

Potential Energy of a Dipole

  • Consider a dipole of dipole moment (\(\vec p\)) in a uniform electric field \(\vec E\).

  • The dipole experiences a torque. 

\(\tau = \vec p \times \vec E\)

  • External work is done in rotating the dipole in opposite direction of the torque. This work is stored as potential energy of dipole.

\(W_{external}=-W_{electric}\)

  • To rotate this dipole from an angle \(\theta\) to \(\phi\), work done by electrical force

\(d\,W_{electric}=-\tau \,d\theta\)

\(d\,W_{electric}=-pE\,sin\,\theta\,d\theta\)

\(\displaystyle\,W_{electric}=-\int \limits ^{\phi}_{\theta}pE\,sin\,\theta\,d\theta\)

\(W_{electric}=-pE\,[-cos\,\theta]^\phi_{\theta}\)

\(W_{electric}=pE\,[cos\,\phi-cos\,\theta]\)

 

  • Change in potential energy

\(\Delta U=-W_{conservative\,force}\)

\(\Delta U=-W_{electric}\)

\(\Delta U=-pE\,[cos\,\phi-cos\,\theta]\)

\(\Rightarrow\;\Delta U=pE\,[cos\,\theta-cos\,\phi]\)

\(\Rightarrow U_\theta-U_\phi=pE[cos\theta-cos\phi]\)

  • From above equation,

\(U=-pE\,cos\,\theta\)

\(U=-\vec p.\vec E\)

Note: Expression of potential energy of dipole is same in uniform and non-uniform electric field.

Illustration Questions

Calculate potential energy of dipole from given figure. Given 60°, – q = – 2 \(\mu\)C,   , q = 2 \(\mu\)C and d = 2 mm, E=1010 V/m. 

A 20 Joule

B –20 Joule

C 200 Joule

D –200 Joule

×

Dipole moment \((\vec p)=q×\vec d\)

\(=2\times10^{-6}\times2\times10^{-3}\)

\( p=4×10^{-9}\) C-m

image

Potential energy is given as,

\(U=-\vec p.\vec E\)

\(U=-(4\times10^{-9})\times(10^{10})\;cos60°\)

\(U=-2\times10\)

\(U=-20\;Joule\)

image

Calculate potential energy of dipole from given figure. Given 60°, – q = – 2 \(\mu\)C,   , q = 2 \(\mu\)C and d = 2 mm, E=1010 V/m. 

image
A

20 Joule

.

B

–20 Joule

C

200 Joule

D

–200 Joule

Option B is Correct

Potential Energy due to Dipole-Dipole Interaction

  • Potential energy of a dipole in the effect of other dipole can be explained as a dipole placed in an electric field.
  • Potential energy of two dipoles placed in an electric field is given as 

\(U=-\vec p.\vec E\)

  • Two dipoles are situated in space in four configuration.

  • Potential energy in any of these configuration is determined as

  • It means potential energy of dipole B is because of electric field of dipole A.
  • Potential energy of any dipole-dipole interaction  is defined as the negative of dot product of dipole moment of dipole itself and the field due to second dipole at a point where first dipole is placed.

\(U=-\vec p. \vec E\)

Illustration Questions

Calculate potential energy of the system. Given \(\vec p_1=2\hat i\) C-m, \(d = 1\,m\) , \(\vec p_2=2\hat i\) C-m

A 3000 Joule

B 145×106 Joule

C –140 Joule

D –144×109 Joule

×

Electric field on the axial line of dipole is given as,

 \(\vec E = \dfrac {2p}{4\pi\epsilon_0r^3} \hat r\)

image

Electric field on dipole 1 due to dipole 2,

 \(E_{1/2}=\dfrac {2\;\vec p_2}{4\pi\epsilon_0r^3}\)

Given, \(\vec p=2\hat i\;\text{C-m}\) , \(\theta = 0°\)

\(\vec E_{1/2}=\dfrac {9\times10^9\times2\times 2\hat i }{(1)^3}\)

\(\vec E_{1/2}=(36\times10^9\;\hat i)\,V/m\)

image image

Electric field on dipole 2 due to dipole 1,

 \(E_{2/1}=\dfrac {2}{4\pi\epsilon_0}\dfrac {\vec p_1}{r^3}\)

Given  \(\vec p_1=2\hat i, \;\theta=0°\)

\(E_{2/1}=\dfrac {9\times10^9\times 2\times 2\hat i}{(1)^3}\)

\(E_{2/1}=36\hat i\times10^9 \;V/m\)

image image

\(U=-\vec p.\vec E = -pE\,cos\theta\)                        \((\theta = 0°)\)

\(U=(-\vec p_1.\vec E_{1/2})+( -\vec p_2. \vec E_{2/1})\)

\(U=(-2\times36\times10^9)+( -2\times36\times10^9)\)

\(U = –144×10^9 \,\,Joule\)

image

Calculate potential energy of the system. Given \(\vec p_1=2\hat i\) C-m, \(d = 1\,m\) , \(\vec p_2=2\hat i\) C-m

image
A

3000 Joule

.

B

145×106 Joule

C

–140 Joule

D

–144×109 Joule

Option D is Correct

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