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.

**Definition-**

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

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

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

Unit:- C-m

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

- Consider a dipole shown in figure.

- 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 \(\ell <<r\)

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

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

- 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 \(\ell <<r\)

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

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

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

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

- 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,

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

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

- 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,

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

A –4.6 Volt

B 1.8 Volt

C 3.9 Volt

D 9.2 Volt

- 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.

A 65.4 Volt

B 60 Volt

C 15 Volt

D 4.5 Volt

- 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

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

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

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

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

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

\(\theta = 90°\)

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

- 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.

- 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.

A 20 Joule

B –20 Joule

C 200 Joule

D –200 Joule

- 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.

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

A 3000 Joule

B 145×106 Joule

C –140 Joule

D –144×109 Joule