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.

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

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.

#### 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
B
C
D

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 $$\ell <<r$$

$$\therefore r^2+\ell^2\simeq r^2$$

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

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

#### 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.

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 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,

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

#### 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

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

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

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

Net field at point A ,

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,

Net field at A,

Net field at A,

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

### 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.

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,

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

#### 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

Taking components of dipole moment

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°

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

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

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

$$= 1.8 \,Volt$$

$$(V_P)_{sin\;60°_A}=0$$

Total potential = 1.8 + 0

= 1.8 Volt

### 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

Option B 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.

#### 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

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

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

$$= 2\times 10^{-9}$$

$$r = 2\,m$$

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

$$V_P=4.5\;\text{ Volt}$$

### 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

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.

#### 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,

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

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

$$\tau =2\times10^{-1}$$

$$\tau =0.2\,\text{N-m}$$

### 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

Option D 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.

#### 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

Potential energy is given as,

$$U=-\vec p.\vec E$$

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

$$U=-2\times10$$

$$U=-20\;Joule$$

### 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

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.

$$U=-\vec p. \vec E$$

#### 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,

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

Electric field on dipole 1 due to dipole 2,

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

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

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

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

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  $$\vec p_1=2\hat i, \;\theta=0°$$

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

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

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

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

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

$$U = –144×10^9 \,\,Joule$$

### 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

Option D is Correct