Chapter 8

Electric Dipoles

It requires a lot of energy to produce a net charge, so most objects do not have a net charge. All objects howevercontain charge and often the centers of positive and negative charge in an object are at different locations. Thebehavior and shape of the electric field of these systems is determined by their electric dipole moment.

8.1 Behavior of Electric Dipoles

This section covers skills and problems involved in drawing dipole fields, deducing the direction of dipole moments fromfields, and predicting the behavior of dipoles in the field of other charge. When drawing field maps for systems withnon-zero total charge, we use the fact that, far from a distribution of charges with non-zero net charge, the electricfield is radial. What happens when the total charge of the distribution is zero? Let’s draw it. Using eight stubs percharge, the field for equal and opposite point charges is drawn below. The dashed circle is the circle at infinity, whichwe have been using for fields with net charge. Note that no lines escape to infinity, which is correct because the systemhas zero net charge. The shape of the field outside the dashed line is the characteristic shape a dipole field. Thestrength of a dipole is given by a vector ~p, the dipole moment.

1

p

Moments of the Electric Field: Any electric field can be expressed as a series of charac-teristic fields whose strength is determined by their “moment”. The net charge of a systemis the system’s monopole moment. The next moment is the dipole moment, defined below.Higher order moments exit: quadrapole, octopole, etc. The long range shape of the field isdetermined by the lowest order non-zero moment. The long range shape of a system withnet charge is radial, determined by its monopole moment. If monopole moment is zero andthe dipole moment non-zero, the long range shape is dipole.

Definition Dipole Moment Vector: The dipole moment vector for a system with zero netcharge, ~p, can be calculated for a collection of charges qi located at the points ~ri using

~p =∑

i

qi~ri

Example 8.1 Dipole Moment of Three ChargesProblem: A 2nC charge is at the origin. Two −1nC charges are at (1cm, 0, 0) and (1cm, 1cm, 0). Calculate thedipole moment vector.

Solution

The dipole moment vector is by definition

~p =∑

i

qi~ri = (2nC)(0, 0, 0) + (−1nC)(1cm, 0, 0) + (−1nC)(1cm, 1cm, 0)

~p = (−2 × 10−11Cm,−1 × 10−11Cm, 0)

Dipole Moment for Equal and Opposite Charges: For a dipole formed of two equal andopposite point charges, the dipole moment points from the negative charge to the positivecharge and has magnitude p = qd where d is the separation between the charges and q isthe charge of the positive charge.

Direction of the Dipole Moment Vector: The dipole moment vector points from thecenter of the negative charge to the center of the positive charge of the charge distribution.

The strength of the dipole, the size of |p|, increases with the amount of charge separated, q, and the amount ofseparation, d, as illustrated below.

Small Amount of Charge Separation - More Charge Separated,

Dipole Moment Larger

Less Separation, Dipole Moment Smaller

Small Dipole Moment

p pp

Far from the charges, all charge distributions with zero total charge but non-zero dipole moment have the charac-teristic dipole electric field. If you see a dipole field, you should be able to draw the dipole moment and tell me thatthe total charge is zero.

pp

The mathematical form of the electric field for a dipole, far from the dipole, is somewhat complicated. We state itfor your reference,

Electric Dipole Field: The electric field of a point dipole at the origin with dipole moment~p is

~E(~r) = k3r(~p · r) − ~p

r3.

This is the field of a point dipole or the field far from a system with zero charge but non-zerodipole moment.

Simplified Electric Dipole Field: The expression above for the dipole field can be simplifiedif a direction for the dipole moment is chosen and only the strength of the field along theaxes is computed. If ~p = py, then along the y axis,

~E(0, y, 0) =2kpy

|y|3.

and along the x-axis

~E(x, 0, 0) = −kpy

|x|3

Notice that the strength of the dipole field falls off as 1/r3 whereas the field of a distribution with net charge falls offas 1/r2. This is why, far from a distribution with net charge, we see only the radial field of a point charge with thetotal charge of the distribution.

Example 8.2 Calculating the Dipole FieldProblem: An electric dipole is formed by equal and opposite point charges with charge ±1nC at ±0.2cmy. Thedipole moment points in the +y direction. Calculate the field at 5m along the x axis and the y axis.

Solution

(a) Calculate the dipole moment: The dipole moment of a simple two charge dipole is p = qd where d = 0.4cmis the separation of the charges p = qd = (1 × 10−9C)(4 × 10−3m) = 4 × 10−12Cm.(b) Calculate the field along the x-axis: The field point at 5m along the x-axis is far from the charges, so theformula for the long range dipole field can be used

~E = −kpy

|x|3= −

(8.99 × 109 Nm2

C2 )(4 × 10−12Cm)y

|5m|3= −2.88 × 10−4 N

Cy

(c) Calculate the field in the y direction: The field point at 5m along the y-axis is far from the charges, so theformula for the long range dipole field can be used

~E =2kpy

|y|3=

2(8.99 × 109 Nm2

C2 )(4 × 10−12Cm)y

|5m|3= 5.75 × 10−4 N

C

8.2 Drawing Dipole Fields

When drawing a dipole electric field map, we need to use a field at infinity that has a dipole shape. The dipole fieldwill arise naturally from our normal process of drawing electric field maps but for some reason everyone scrunches allthe field down very close to the charges. Therefore, to get the correct long range field it helps to draw the dipole fieldat infinity first. So to draw a dipole field, we use the same process as for a field with net charge (monopole), exceptreplace the long range part with the following two steps:

Determine Direction of Dipole Moment: The dipole moment is directed from the centerof negative charge to the center of positive charge. Draw it on your figure.

Draw a Dipole Long Range Field: Draw the circle at infinity and draw a dipole fieldmatching your dipole moment.

Example 8.3 Field of Point DipoleProblem: Draw the electric field of an electric dipole formed of two point charges with dipole moment in the +ydirection.

Solution

(a) Draw the Charges: Draw the electric charges at the given locations to scale.Since we are given an electric dipole along the y-axis, draw equal and oppositecharges along the y-axis.

y

x

(b) Draw the Dipole Moment: The dipole moment vector is drawn from thecenter of negative charge to the center of positive charge. For two point charges,the electric dipole is drawn from the negative to the positive charge.

y

x

p

(c) Draw the Long Range Dipole Field: For a charge distribution that has zero net charge and a non-zero dipolemoment, the electric field far from the charge has the characteristic shape of an electric dipole.

p

(d) Draw Stubs of Field Lines: I chose eight lines per charge. The field lines exit at the positive charge and enterat the negative.

p

(e) Connect the Lines: Connect and smooth the innerand outer lines. Jiggle until you get something appropri-ately symmetric.

p

8.3 Qualitative Dipole Behavior

Systems of charge whose lowest order non-zero moment is the dipole moment behave differently than systems of chargewith net charge. Our model for an electric dipole will be two equal and opposite charges at each end of a stick.

Barbell Model of Dipole: When considering the motion ofdipoles, we will model them using equal and opposite point chargeson a stick, as shown to the right. +

_ p

Our barbell dipole is placed in a number of electric fields below. The force on each charge, ~F+ and ~F−, and the net

force, ~Fnet = ~F+ + ~F−, on the dipole is drawn in each case.

Figure (a) Equilibrium

+

_

Zero Net Force

Figure (b) Away from Equilibrium Figure (c) Non-Uniform Field

+

_

Non-Zero Net Force Zero Net Force

+

_

pp p

F+

F−

F+

F−

F+

F−

Fnet

In figure (a) the field is uniform and the dipole moment ~p aligns with the field. The net force is zero and the forceson the dipole do not tend to rotate the dipole. This is the equilibrium position of the dipole. In figure (b), the dipole

is rotated away from equilibrium. The net force is still zero, but the force on each charge tends to rotate the dipoletoward equilibrium.

Dipoles Rotate to Align with Field: A dipole placed in an electric field is at equilibriumwhen the dipole moment points in the same direction as the field line. A dipole that is notat equilibrium will tend to rotate toward alignment with the field line.

Dipoles In a Uniform Field Feel Zero Net Force: If a dipole is placed in a uniform electricfield, constant through space, then the total force (but not the torque) is zero since the forceson the plus and minus charge are equal and opposite. So the dipole rotates but its center ofmass stays in the same place.

In figure (c), the field is not uniform. The positive and negative charges forming the dipole experience forces of differentmagnitudes and directions and therefore there is a net force on dipole.

Net Force on Dipoles In a Non-Uniform Field: If a dipole is placed in a non-uniformfield, the two charges experience difference forces, and the direction of the net force must bedetermined by adding these forces.

Example 8.4 Electric Dipole in Uniform FieldProblem: A uniform electric field is directed in the +x direction. A barbell dipole with dipole moment direction inthe +y direction is placed in the field.

(a)Draw the field and the barbell dipole.

(b)Draw the electric force vectors on the charges at the ends of the dipole.

(c)Indicate the direction of rotation of the dipole.

Solution to Part (a)

The field lines are evenly spaced since we are told the field is uniform. The dipole moment, in the +y direction here,always points from the negative to the positive charge. See figure.

Solution to Part (b)

The force on the positive charge will point the same direction as the field; the force on the negative charge will pointin the opposite direction. The forces have the same magnitude since the field is uniform. See figure.

Solution to Part (c)

The dipole will rotate in the clockwise direction based on the forcesdrawn. The dipole moment will tend to align itself with the field lines. y

x

+

−

F+

F−

p Rotation

8.4 Dipole Mechanics

We argued in previous section that a dipole will rotate to align with an uniform electric field, but feel no net force.Since the dipole tends to rotate, it must experience a net torque. If the field is not uniform, the dipole will experiencea net force. This means the force depends on how the field changes. If the dipole is allowed to rotate it will come toequilibrium (if there are losses in the system) with its dipole moment aligned with the field. This behavior implies thedipole is seeking the minimum in some potential energy function. So to quantitatively describe the mechanics of anelectric dipole, we need to evaluate the torque, net force, and potential energy.

8.4.1 Potential Energy of a Dipole in an Electric Field

To calculate the potential energy of an electric dipole with orientation θ with respect to a uniform field, we have tocalculate the work required to rotate the dipole from the location of zero potential energy to the orientation θ. Infigure (a) below, the dipole is drawn in its minimum energy orientation, aligned with the field. In figure (b), the dipolehas been rotated an angle θ away from equilibrium. In both figures, the dipole moment vector, ~p, is drawn.

Figure (a) Minimum Energy Figure (b)

+

_

d

+

_

∆h

∆h

θ

p

p

The difference in potential energy, ∆U , from figure (a) to figure (b) is the work, W , an external agent would haveto do to rotate the dipole. Work is force times the distance in the direction of the force. Both the positive and negativecharge moved a distance ∆h against the force of field. The work done is W = F+∆h+F−∆h = 2qE∆h, where F+ isthe force on the positive charge, q is the magnitude of the positive charge, and E is the electric field. If θ is the anglebetween the dipole moment vector, ~p, and the field ~E, and d is the length of the dipole, then ∆h = d/2 − d/2 cos θ.Substituting gives the change in potential energy to rotate from figure (a) to figure (b).

∆U = 2qE

(

d

2−

d

2cos θ

)

= −pE cos θ − pE

where I have used |~p| = qd. It is customary to choose the zero of potential energy so the pE goes away.

Potential Energy of an Electric Dipole: The potential energy U of an electric dipole withdipole moment ~p in a uniform electric field ~E is

U = −pE cos θ = −~p · ~E

where θ is the angle between the dipole moment and the field. The second expression usesthe vector dot product, which we will review in Chapter 11.

8.4.2 Torque on an Electric Dipole

An electric field tends to make an electric dipole rotate, and therefore exerts a torque on the dipole. An electric dipolein a uniform electric field ~E is drawn below. The forces on each charge are also drawn.

+

_

Moment Arm

p

θ

F+

F−

From UPI, the torque τ exerted on the object is the force multiplied by the moment arm, the perpendicular distanceto the line of action of the force. The moment arm and lines of action are drawn above. The torque will be calculatedabout the center of the dipole. The torque on the dipole in the figure causes it to rotate in the counterclockwisedirection. Both forces, ~F+ and ~F−, exert a torque on the object. The total torque is the sum of the torques of thetwo forces, τ = τ+ + τ−. The length of the moment arm is (d/2) sin θ for both forces if the separation of the chargesis d and the angle θ is measured from the dipole moment to the field. The angle θ is positive above. The force oneach charge is qE. The total torque is then τ = qEd sin θ or τ = pE sin θ where I have used |~p| = qd.

Torque on an Electric Dipole: The torque, τ , on an electric dipole with dipole momentvector ~p in an electric field ~E is

τ = pE sin θ

where θ is measured from ~p to ~E. A positive torque causes a counterclockwise angularacceleration.

When we reach magnetic dipoles and have some experience with the vector cross product, the above expression willbe re-written as ~τ = ~p × ~E.

8.4.3 Force on an Electric Dipole in a Non-Uniform Field

We have already argued that a uniform electric field exerts zero netforce on an electric dipole. An electric dipole is drawn below in a non-uniform electric field that points generally in the y direction at thedipole. The force on the charges forming the dipole are drawn as wellas the net force, ~Fnet. The x-component is an artifact of how large Ihave drawn the dipole. As d, the length of the dipole, gets smaller thex component vanishes.

+

_ y

x

θ F+

F−

Fnet

p

We would like to estimate the force on the dipole in the limit the length of the dipole, d, is small. If the field points

generally in the y direction at the location of the dipole then at the dipole we can write the field ~E = E(y)y. The netforce will point generally in the y direction and have magnitude, Fnet = qE(y+)− qE(y−) where y+ is the location ofthe + charge and y− is the location of the − charge. If the separation of charges d is small, then this is approximately

Fnet = qdE

dy(y+ − y−) = q

dE

dy(d cos θ) = p

dE

dy(cos θ)

where θ is the angle between the dipole moment and the field and once again I have use |~p| = dq.

Force on an Electric Dipole in a Non-Uniform Field: The net force on an electric dipolewith dipole moment ~p in an electric field the points in the y direction, ~E = E(y)y, is

Fnet = pdE

dy(cos θ)

where θ is the angle between the dipole moment vector and the y axis.

Note, if the dipole moment aligns with the field (θ = 0), the dipole feels a force toward stronger field. If the dipoleanti-aligns with the field (θ = 180◦), the dipole feels a force toward weaker field.

Example 8.5 Rotation of a Water MoleculeProblem: The NIST database gives the dipole moment of water as p = 1.85debye = 6.18× 10−30Cm. As you workthrough these databases the profusion of different systems of units is really annoying. A water molecule is placed inthe electric field of the golf tube modelled as an infinite line of charge along the z axis. The golf tube has linear chargedensity λ = −0.10µC/m. The molecule is 4cm from the axis of the tube along the x axis. The angle between thedipole moment of the molecule and the electric field is θ = 45◦.

(a)Calculate the potential energy of the molecule.

(b)Calculate the torque exerted on the molecule by the field.

(c)Calculate the net force on the molecule.

Solution to Part (a)

The electric field of the golf tube at the water molecule is

E =λ

2πε0d=

−0.1 × 10−6C/m

2π(8.85 × 10−12 C2

Nm2 )(0.04m)= −45000

N

C

The potential energy of an electric dipole in an electric field is

U = −pE cos θ = −(6.18×10−30Cm)(−45000) cos 135◦ = 2.0×10−25J

where θ = 135◦ is the angle between the dipole and the field. Thisdrawing is way out of scale, a molecule is tiny, so we can pretend E isin the same direction at either end of the dipole.

_

+θ

α

p

Solution to Part (b)

The magnitude of the torque on the water molecule is

|τ | = pE sinα = |(6.18 × 10−30Cm)(−45000N

C) sin 45◦| = 2.0 × 10−25Nm

where α is the angle between the dipole moment and the x axis. Notice that torque and energy have the same units1 Joule = 1Nm.

Solution to Part (c)

The force on the dipole is

F = pdE

drcos θ

The derivative of the electric field isdE

dr=

d

dr

λ

2πε0r= −

λ

2πε0r2

The force is then

F = pdE

drcos θ = −

pλ

2πε0r2cos θ = −

(6.18 × 10−30Cm)(−0.1 × 10−6C/m)

2π(8.85 × 10−12 C2

Nm2 )(0.04m)2cos 45◦ = 4.9 × 10−24N

The positive sign indicates the force is outward from the golf tube.