Dipole in a Uniform External Field

Electric Charges and Fields · CBSE

Theory

The set-up

A permanent dipole of moment $\vec{p}$ — permanent meaning $\vec{p}$ exists whether or not the field is applied, it is not induced by the field — is placed in a uniform external field $\vec{E}$ , with $\vec{p}$ making an angle $\theta$ with $\vec{E}$ .

Derivation of the torque — the standard 2- or 3-mark answer

The charge $+q$ experiences a force $q\vec{E}$ along $\vec{E}$ ; the charge $-q$ experiences $-q\vec{E}$ , equal in magnitude and opposite in direction.
Net force $= q\vec{E}+(-q\vec{E}) = 0$ , because the field is the same at both charges. The dipole does not translate.
But the two forces do not act along the same line. They are equal, antiparallel and non-collinear, so they form a couple, and produce a torque. Since the net force is zero, the torque is the same about every origin.
Torque = magnitude of either force $\times$ the perpendicular distance between their lines of action (the arm of the couple). With the dipole of length $2a$ inclined at $\theta$ , the arm is $2a\sin\theta$ :

$\tau = qE\times 2a\sin\theta = (q\cdot 2a)E\sin\theta = pE\sin\theta$

In vector form $\vec{\tau} = \vec{p}\times\vec{E}$ , since $\lvert\vec{p}\times\vec{E}\rvert = pE\sin\theta$ and the direction of the cross product is perpendicular to the plane containing $\vec{p}$ and $\vec{E}$ , along the outward normal to that plane.
Definition to quote: the torque on a dipole in a uniform field is the moment of the couple formed by the two equal and opposite forces acting on its charges, of magnitude $pE\sin\theta$ , directed perpendicular to both $\vec{p}$ and $\vec{E}$ .
The torque always tends to rotate the dipole into alignment with $\vec{E}$ ; when $\vec{p}$ is aligned with $\vec{E}$ the torque vanishes.

The diagram in words

Draw $\vec{E}$ as a set of equally spaced parallel horizontal arrows. Place the dipole at an angle $\theta$ to them, $-q$ at the lower left, $+q$ at the upper right, separation $2a$ , with $\vec{p}$ drawn from $-q$ to $+q$ . At $+q$ draw a force $q\vec{E}$ to the right, at $-q$ a force $q\vec{E}$ to the left. Drop a perpendicular between the two lines of action and mark it $2a\sin\theta$ — that is the arm of the couple. Show $\vec{\tau}$ coming out of the plane of the paper.

Special orientations

Angle $\theta$ between $\vec{p}$ and $\vec{E}$	Torque	Potential energy	Equilibrium
$0^\circ$	$0$	$-pE$ , minimum	stable
$90^\circ$	$pE$ , maximum	$0$	none
$180^\circ$	$0$	$+pE$ , maximum	unstable

Stable equilibrium: $\vec{p}$ parallel to $\vec{E}$ . The torque is zero, and any small angular displacement produces a restoring torque that brings the dipole back.
Unstable equilibrium: $\vec{p}$ antiparallel to $\vec{E}$ . The torque is again zero, but any small displacement produces a torque that turns the dipole further away.
Depicting them: for stable equilibrium draw the dipole arrow lying along the field arrows, $-q$ upstream and $+q$ downstream; for unstable equilibrium draw it reversed, $+q$ upstream.

Potential energy

$U = -\vec{p}\cdot\vec{E} = -pE\cos\theta$

A dipole of charges $+q$ and $-q$ separated by $L$ in stable equilibrium therefore has $U = -pE = -qLE$ .

Perpendicular vectors in the torque expression

Because $\vec{\tau} = \vec{p}\times\vec{E}$ is a cross product, it is perpendicular to both factors. The two pairs of mutually perpendicular vectors are $(\vec{\tau},\vec{p})$ and $(\vec{\tau},\vec{E})$ . Note that $\vec{p}$ and $\vec{E}$ are in general not perpendicular to each other.

Non-uniform field

If the field is not uniform, the forces on $+q$ and $-q$ are no longer equal in magnitude, because the two charges sit where the field strengths differ. The net force is then non-zero, and there is in general a torque as well.
Consequently a dipole in a non-uniform field undergoes both translational and rotational motion.
The simple cases: if $\vec{p}$ is parallel to $\vec{E}$ the net torque is zero, but the dipole is pushed in the direction of increasing field. If $\vec{p}$ is antiparallel to $\vec{E}$ the torque is again zero, but the net force is in the direction of decreasing field — that is, opposite to $\vec{p}$ when the field increases along $\vec{p}$ .
The assertion-reason version: a dipole in a non-uniform field has translatory as well as rotatory motion, and the reason — it experiences both a force and a torque — is the correct explanation.
Everyday application: a comb run through dry hair picks up charge and attracts small pieces of paper. The paper is not charged, but the comb's field polarises it, inducing a dipole moment along the field. Because the comb's field is non-uniform, this induced dipole experiences a net force towards the comb — which is why the paper moves towards it.

Resultant of two dipoles

For two dipoles of moments $p_1$ and $p_2$ inclined at $\alpha$ ,

$p_r = \sqrt{p_1^2+p_2^2+2p_1p_2\cos\alpha}$

For two identical dipoles $p$ at $120^\circ$ : $p_r = \sqrt{2p^2+2p^2\cos120^\circ} = \sqrt{2p^2-p^2} = p$ , and the resultant bisects the angle between them. If the two dipole axes are placed symmetrically so the resultant makes $30^\circ$ with $\vec{E}$ (along the $x$ -axis), the net torque has magnitude $\tau = p_rE\sin30^\circ = \dfrac{1}{2}pE$ , directed along the negative $z$ -axis.

Numerical pattern

A dipole of length $2\,\text{cm}$ at $30^\circ$ to a field of $2\times10^{5}\ \text{N C}^{-1}$ experiences a torque of $8\times10^{-3}\ \text{N m}$ . From $\tau = pE\sin\theta$ ,

$p = \dfrac{8\times10^{-3}}{2\times10^{5}\times0.5} = 8\times10^{-8}\ \text{C m}$

and since $p = q\times2l$ with $2l = 2\times10^{-2}\,\text{m}$ , $q = 4\times10^{-6}\,\text{C} = 4\,\mu\text{C}$ .

Standard results to carry into the exam

Uniform field: net force zero, torque $pE\sin\theta$ , motion purely rotational.
Non-uniform field: net force and torque both non-zero, motion translational plus rotational.
Maximum torque $pE$ at $\theta = 90^\circ$ ; zero torque at $0^\circ$ and $180^\circ$ .
Work done in rotating a dipole from $\theta_1$ to $\theta_2$ in a uniform field is $W = pE(\cos\theta_1-\cos\theta_2)$ , which is also the change in potential energy.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
(a) Derive an expression for the electric field $E$ due to a dipole of length ' $2a$ ' at a point distant $r$ from the centre of the dipole on the axial line. (b) Draw a graph of $E$ versus $r$ for $r \gg a$ .
5 marksCBSE Class 12 Physics — Long Answer · 3/5, AI 2017, An · Section E, 5 marksAnswer 1 ↓
2
An electric dipole of dipole moment $\vec{p}$ consists of point charges $+q$ and $-q$ separated by a distance $2a$ apart. Deduce the expression for the electric field $\vec{E}$ due to the dipole at a distance $x$ from the centre of the dipole on its axial line, in terms of the dipole moment $\vec{p}$ . Hence show that in the limit $x \gg a$ , $\vec{E} \to 2\vec{p}/(4\pi\varepsilon_0 x^3)$ .
5 marksCBSE Class 12 Physics — Long Answer · 3/5, Delhi 2015, Ev · Section E, 5 marksAnswer 2 ↓
3
Write the expression for the torque $\vec{\tau}$ acting on a dipole of dipole moment $\vec{p}$ placed in an electric field $\vec{E}$ .
1 markCBSE Class 12 Physics — Very Short Answer · Foreign 2015 · Section A, 1 markAnswer 3 ↓
4
Derive the expression for the torque acting on an electric dipole, when it is held in a uniform electric field. Identify the orientation of the dipole in the electric field, in which it attains a stable equilibrium.
2 marksCBSE Class 12 Physics — SA · 2020 · Section B, 2 marksAnswer 4 ↓
5
Two small identical electric dipoles $AB$ and $CD$ , each of dipole moment $\vec{p}$ , are kept at an angle of $120°$ to each other in an external electric field $\vec{E}$ pointing along the $x$ -axis (with $A$ at $+q$ along $+y$ , $B$ at $-q$ ; $C$ at $+q$ , $D$ at $-q$ , the two dipole axes at $120°$ to each other). Find (a) the dipole moment of the arrangement, and (b) the magnitude and direction of the net torque acting on it.
3 marksCBSE Class 12 Physics — SA · AI 2020 · Section C, 3 marksAnswer 5 ↓
6
Depict the orientation of the dipole in (a) stable, (b) unstable equilibrium in a uniform electric field.
1 markCBSE Class 12 Physics — Very Short Answer · 1/3, Delhi 2017, R · Section A, 1 markAnswer 6 ↓
7
(a) Obtain the expression for the torque $\vec{\tau}$ experienced by an electric dipole of dipole moment $\vec{p}$ in a uniform electric field $\vec{E}$ . (b) What will happen if the field were not uniform?
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2017 · Section E, 5 marksAnswer 7 ↓
8
An electric dipole of dipole moment $\vec{p}$ is placed in a uniform electric field $\vec{E}$ . Obtain the expression for the torque $\vec{\tau}$ experienced by the dipole. Identify two pairs of perpendicular vectors in the expression.
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2015C, Cr · Section E, 5 marksAnswer 8 ↓
9
(a) Define torque acting on a dipole of dipole moment $\vec{p}$ placed in a uniform electric field $\vec{E}$ . Express it in vector form and point out the direction along which it acts. (b) What happens if the field is non-uniform? (c) What would happen if the external field $\vec{E}$ is increasing (i) parallel to $\vec{p}$ and (ii) anti-parallel to $\vec{p}$ ?
5 marksCBSE Class 12 Physics — Long Answer · Foreign 2016 · Section E, 5 marksAnswer 9 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

10
An electric dipole of length $2\,\text{cm}$ is placed at an angle of $30°$ with an electric field $2\times10^5\,\text{N/C}$ . If the dipole experiences a torque of $8\times10^{-3}\,\text{Nm}$ , the magnitude of either charge of the dipole is
- (a)
  $4\,\mu\text{C}$
- (b)
  $7\,\mu\text{C}$
- (c)
  $8\,\text{mC}$
- (d)
  $2\,\text{mC}$
1 markCBSE Class 12 Physics — Multiple Choice · 2023 · Section A, 1 markAnswer 10 ↓
11
An electric dipole placed in a non-uniform electric field can experience
- (a)
  a force but not a torque
- (b)
  a torque but not a force
- (c)
  always a force and a torque
- (d)
  neither a force nor a torque
1 markCBSE Class 12 Physics — Multiple Choice · 2020, R · Section A, 1 markAnswer 11 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
(a) Consider a dipole with charges $+q$ (at distance $r-a$ ) and $-q$ (at distance $r+a$ ) from an axial point $P$ , with dipole moment $\vec{p}$ pointing from $-q$ to $+q$ . Field due to $-q$ at $P$ : $\vec{E}_{-q} = \dfrac{-q}{4\pi\varepsilon_0(r+a)^2}\hat{p}$ (towards $-q$ , i.e. left) Field due to $+q$ at $P$ : $\vec{E}_{+q} = \dfrac{q}{4\pi\varepsilon_0(r-a)^2}\hat{p}$ (towards right) Resultant field: $\vec{E}_{axial} = \vec{E}_{+q}+\vec{E}_{-q} = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{(r-a)^2}-\dfrac{1}{(r+a)^2}\right]\hat{p}$ $= \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{4ar}{(r^2-a^2)^2}\hat{p} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2pr}{(r^2-a^2)^2}\hat{p}$ (using $p=q\cdot2a$ ) For $r \gg a$ , $a^2$ can be neglected compared to $r^2$ : $\vec{E}_{axial} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2\vec{p}}{r^3}$ (directed along $+\hat{p}$ ) (b) $E$ falls off as $1/r^3$ : a steep monotonically decreasing curve of $E$ against $r$ , approaching zero as $r$ increases.
2.↑ back to the question
(Same derivation as Q28, with $r$ replaced by $x$ .) Field at an axial point at distance $x$ from the centre: $\vec{E}_{axial} = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{(x-a)^2}-\dfrac{1}{(x+a)^2}\right]\hat{p} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2px}{(x^2-a^2)^2}\hat{p}$ For $x \gg a$ , $a^2$ is negligible compared to $x^2$ : $\vec{E} \to \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2\vec{p}}{x^3}$ — as required.
3.↑ back to the question
The torque acting on a dipole of dipole moment $\vec{p}$ placed in an electric field $\vec{E}$ is given by $\vec{\tau} = \vec{p}\times\vec{E}$ , i.e., $\tau = pE\sin\theta$ , where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$ .
4.↑ back to the question
When an electric dipole ( $+q$ at $A$ , $-q$ at $B$ , separation $2a$ ) is placed in a uniform field $\vec{E}$ making angle $\theta$ with $\vec{p}$ , the two charges experience equal and opposite forces $qE$ , which cancel — so the net force is zero. However, these forces are not collinear, so they constitute a couple. Torque $=$ magnitude of either force $\times$ perpendicular distance between the lines of action $= qE\times(2a\sin\theta) = (q\cdot2a)E\sin\theta$ $\therefore\ \tau = pE\sin\theta$ , or in vector form, $\vec{\tau} = \vec{p}\times\vec{E}$ Stable equilibrium: when $\theta = 0°$ (i.e., $\vec{p}$ parallel to $\vec{E}$ ), $\tau = 0$ and any small angular displacement produces a restoring torque bringing the dipole back — this is the position of stable equilibrium.
5.↑ back to the question
(a) The resultant dipole moment of the two dipoles (each of magnitude $p$ , at $120°$ to each other) is $p_r = \sqrt{p^2+p^2+2p\cdot p\cos120°} = \sqrt{2p^2+2p^2\left(-\dfrac{1}{2}\right)} = \sqrt{2p^2-p^2} = p$ The resultant dipole moment makes an angle of $60°$ with the $y$ -axis (i.e., $30°$ with the $x$ -axis). (b) Torque, $\tau = p_rE\sin\theta$ , where $\theta = 30°$ is the angle between the resultant dipole moment and $\vec{E}$ (the $x$ -axis): $\tau = pE\sin30° = \dfrac{1}{2}pE$ The direction of the torque is along the negative $z$ -direction.
6.↑ back to the question
(a) Stable equilibrium: $\theta = 0°$ , i.e., $\vec{p}$ parallel to $\vec{E}$ (dipole aligned along the field) — here $\tau=0$ and any small displacement is restored. (b) Unstable equilibrium: $\theta = 180°$ , i.e., $\vec{p}$ anti-parallel to $\vec{E}$ (dipole aligned opposite to the field) — here also $\tau=0$ , but any small displacement grows rather than being restored.
7.↑ back to the question
(a) (Same derivation as Q33.) When a dipole is placed in a uniform field, the two equal and opposite forces $qE$ on its charges cancel (net force zero), but being non-collinear they produce a torque: $\tau = Fr_\perp = qE\cdot2a\sin\theta = (q\cdot2a)E\sin\theta$ , i.e., $\tau = pE\sin\theta$ , or $\vec{\tau}=\vec{p}\times\vec{E}$ — this torque tends to align the dipole along the field direction. (b) If the field is not uniform, the forces on the two charges no longer cancel exactly, so the dipole experiences a net force in addition to the torque, and undergoes both translational and rotational motion.
8.↑ back to the question
(Same derivation as Q33/36.) $\tau = pE\sin\theta$ , or in vector form, $\vec{\tau} = \vec{p}\times\vec{E}$ . Since $\vec{\tau}$ is the cross product of $\vec{p}$ and $\vec{E}$ , it is perpendicular to both. The two pairs of perpendicular vectors are: (i) $(\vec{\tau},\vec{p})$ and (ii) $(\vec{\tau},\vec{E})$ .
9.↑ back to the question
(a) When a dipole of moment $\vec{p}$ , making angle $\theta$ with $\vec{E}$ , is placed in a uniform field, the torque acting on it is $\tau = pE\sin\theta$ , or $\vec{\tau} = \vec{p}\times\vec{E}$ ; its direction is perpendicular to both $\vec{p}$ and $\vec{E}$ . (b) If the field is non-uniform, there is a net force on the dipole in addition to the torque, and the resulting motion is a combination of translation and rotation. (c) (i) If $\vec{E}$ is increasing parallel to $\vec{p}$ , then $\theta = 0°$ , so the torque remains zero, but the net force on the dipole is in the direction of increasing field — hence it undergoes linear motion along the direction of $\vec{p}$ . (ii) If $\vec{E}$ is increasing anti-parallel to $\vec{p}$ , the torque is again zero, but the net force is in the direction of the increasing field, which is opposite to $\vec{p}$ — hence linear motion opposite to the dipole moment.
10.Answer: (A)↑ back to the question
Given: $2l = 2\,\text{cm}$ , $\theta=30°$ , $E=2\times10^5\,\text{N/C}$ , $\tau = 8\times10^{-3}\,\text{Nm}$ $\tau = pE\sin\theta \ \Rightarrow\ 8\times10^{-3} = p\times2\times10^5\times\sin30°$ $p = 8\times10^{-8}\,\text{C m}$ Since $p = q\times2l$ : $8\times10^{-8} = q\times2\times10^{-2} \ \Rightarrow\ q = 4\,\mu\text{C}$
11.Answer: (C)↑ back to the question
Correct option: (c). An electric dipole placed in a non-uniform electric field always experiences both a torque and a net force (since the forces on the two charges, being at slightly different field strengths, do not cancel).