Electric Dipole: Difference between revisions

From Physics Book
Jump to navigation Jump to search
← Older edit
Jmorton32 (talk | contribs)
35 edits
No edit summary
 
(109 intermediate revisions by 14 users not shown)
Line 1: Line 1:
An '''Electric Dipole''' is a pair of equal and opposite [[Point Charge]]s separated by a small distance.
'''Claimed By: Yashwin Thammiraju, Spring 2026'''


claimed by [[User:Jmorton32|Jmorton32]] ([[User talk:Jmorton32|talk]]) 02:52, 19 October 2015 (EDT)
==Summary==
 
An electric dipole is made up of two point charges that have equal but opposite electric charges (q) and are separated by a short distance (d).
 
[[File:dipo.jpg|An Electric Dipole]]
 
The electric field of a dipole is inversely proportional to the cube of the distance from the dipole (<math>1/r^3</math>), unlike a single point charge which falls off by <math>1/r^2</math>. The field's magnitude and direction are highly dependent on whether you are observing it along the parallel axis (the line separating the two charges) or the perpendicular axis (the bisector).
 
A temporary dipole can be created when you place a neutral atom in an external electric field. Due to the movement of the electron cloud relative to the nucleus, the atom polarizes (shifting negative charge to one side and positive charge to the other), yielding a separation of charge.
 
Electric dipoles are characterized by their '''dipole moment''' (<math>\vec{p}</math>), a vector quantity measuring the strength and separation of the positive and negative electrical charges within a system. For two point charges, +q and -q, separated by a distance d, the magnitude of the dipole moment is:
 
<math>p = qd</math>
 
A prime example of a permanent dipole in nature is the water molecule (H<sub>2</sub>O), which forms a 105-degree angle between the two hydrogen atoms connected to the oxygen. Because oxygen has a greater electronegativity, it pulls more strongly on the shared electrons. Consequently, the oxygen end of the molecule becomes more negatively charged compared to the hydrogen end, and the net electric dipole moment points towards the oxygen atom.
 
[[File:Water.png|300px|thumb|Dipole moment of water]]
 
==Computational Model==
To better visualize the electric field generated by an electric dipole, we can use a computational model. The GlowScript simulation below calculates the exact superposition of the point charges and displays the resulting electric field vectors at various observation locations.
 
<html><iframe src="https://trinket.io/embed/glowscript/31d0f9ad9e" width="100%" height="400" frameborder="0" marginwidth="0" marginheight="0" allowfullscreen></iframe></html>


==Mathematical Models==
==Mathematical Models==


===An Exact Model===
[[File:Phys2212 dipole image.PNG|300px|thumb|Polarization by an electric field]]
An electric dipole is constructed from two point charges: one at position <math>[\frac{d}{2}, 0]</math> and one at position <math>[\frac{-d}{2}, 0]</math>. These point charges have equal and opposite charge magnitudes. We wish to calculate the exact electric field due to the dipole at some observation point P in the plane (see the figure). Point P can be defined by its coordinates <math>[p_x, p_y]</math> from the midpoint of the dipole, or by a distance r and an angle <math>\theta</math>.
Using the superposition principle, the net electric field at P is <math>E_{net} = E_{q_+} + E_{q_-}</math>. We can break this down into x and y components:
* <math>E_{net_x} = E_{q_{+x}} + E_{q_{-x}}</math>
* <math>E_{net_y} = E_{q_{+y}} + E_{q_{-y}}</math>
Let <math>\theta_+</math> be the angle from q<sub>+</sub> to P. The y-component of the positive charge's field is <math>E_{q_{+y}} = E_{q_+} \sin(\theta_+)</math>.
To find <math>\theta_+</math> and its counterpart <math>\theta_-</math>, we look at the geometry. <math>\theta_+</math> belongs to a right triangle with an opposite side length of <math>p_y</math> and an adjacent side length of <math>p_x - \frac{d}{2}</math>. Therefore:
<math>\sin(\theta_+) = \frac{p_y}{\sqrt{(p_x - \frac{d}{2})^2+p_y^2}}</math>
The denominator here represents the hypotenuse, which is simply the distance <math>|\vec r_+|</math> from the positive charge to the observation point. A similar geometric analysis for the negative charge gives:
<math>\sin(\theta_-) = \frac{p_y}{\sqrt{(p_x + \frac{d}{2})^2+p_y^2}}</math> where the denominator is <math>|\vec r_-|</math>.


===An Exact Model===
The general formula for the magnitude of an electric field from a point charge is <math>|E| = \frac{1}{4\pi\epsilon_0} \frac{q}{|\vec r|^2}</math>. Applying this to both charges:
[[File:Dipole.png|300px|thumb|An Electric Dipole]]
<math>E_{net_y} = \frac{1}{4\pi\epsilon_0} \frac{q_+}{|\vec r_+|^2} \sin(\theta_+) + \frac{1}{4\pi\epsilon_0} \frac{q_-}{|\vec r_-|^2} \sin(\theta_-)</math>
Since an electric dipole is made up of 2 electric point charges, the electric field of the dipole can be calculated by summing the electric fields contributed by each point charge.  In the example, the field at point P, <math>E_{P}</math> is equivalent to the sum of the electric field from the positive charge <math>q+</math> and the negative charge <math>q-</math>.  
In other words, <math>E_{P} = E_{P_{q+}} + E_{P_{q-}}</math>.


Substituting in the equation for the electric field from a point charge we get <math>\vec E_{P} = \frac{1}{4\pi\epsilon_{0}} \times \frac{q_{+}}{|r_{+}|^{2}} \times \hat r_{+} + \frac{1}{4\pi\epsilon_{0}} \times \frac{q_{-}}{|r_{-}|^{2}} \times \hat r_{-}</math>.
Noting that <math> q_+ = -q_-</math>, we can factor out the charge:
<math>E_{net_y} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{1}{|\vec r_+|^2}\sin(\theta_+) - \frac{1}{|\vec r_-|^2}\sin(\theta_-)\Bigg)</math>


A simple refactoring gives <math>\vec E_{P} = \frac{1}{4\pi\epsilon_{0}} \times (\frac{q_{+}}{|r_{+}|^{2}}\hat r_{+} + \frac{q_{-}}{|r_{-}|^{2}} \hat r_{-})</math>.
Substituting our expanded radii and sines into the equation yields:
<math>E_{net_y} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{1}{(p_x - \frac{d}{2})^2 +p_y^2 } \frac{p_y}{\sqrt{(p_x - \frac{d}{2})^2+p_y^2}} - \frac{1}{(p_x + \frac{d}{2})^2 +p_y^2 } \frac{p_y}{\sqrt{(p_x + \frac{d}{2})^2+p_y^2}} \Bigg)</math>


Since <math>q_+ = -1 * q_-</math>, this is equivalent to <math>\vec E_{P} = \frac{q_+}{4\pi\epsilon_{0}} \times (\frac{\hat r_{+}}{|r_{+}|^{2}} - \frac{\hat r_{-}}{|r_{-}|^{2}})</math>
Combining the denominators simplifies this to the exact analytical form for the y-component:
<math>E_{net_y} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{p_y}{\Big((p_x - \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2} } - \frac{p_y}{\Big((p_x + \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2} } \Bigg)</math>


<math>|r_+| \text{ and } |r_-|</math> can then be calculated as <math>|r_+| = \sqrt{(|r_x| + \frac{d}{2})^2 + |r_y|^2}</math>, by decomposing <math>\vec r</math> into its components and factoring.
The derivation for the x-direction follows the exact same logic, using cosine (adjacent over hypotenuse) instead of sine. The result is:
<math>E_{net_x} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{p_x - \frac{d}{2}}{\Big((p_x - \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2} } - \frac{p_x + \frac{d}{2}}{\Big((p_x + \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2} } \Bigg)</math>


By a similar method, <math>|r_-| = \sqrt{(|r_x| - \frac{d}{2})^2 + |r_y|^2}</math>.
These formulae provide the exact electric field due to an electric dipole anywhere on the 2D plane.


By substituting <math>|\vec r| \cos(\theta) \text{ and } |\vec r| \sin(\theta)</math> for  <math>|r_x| \text{ and } |r_y|</math> respectively, we get
==Special Cases (Approximations)==
When the observation distance is much greater than the separation distance (<math>r \gg d</math>), we can simplify the exact models into the standard dipole approximations. Let <math>a = \frac{d}{2}</math>.


<math>|r_+| = \sqrt{(|\vec r| \cos(\theta) + \frac{d}{2})^2 + (|\vec r| \sin(\theta))^2}</math> and <math>|r_-| = \sqrt{(|\vec r| \cos(\theta) - \frac{d}{2})^2 + (|\vec r| \sin(\theta))^2}</math>
===On the Parallel Axis===
On the axis running through the two charges, <math>p_y = 0</math>, meaning <math>E_{net_y} = 0</math>. Plugging <math>p_y = 0</math> into our exact <math>E_{net_x}</math> formula:


We next expand the squared terms inside the radical: <math>\sqrt{(|\vec r| \cos(\theta))^2 + \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4} +(|\vec r| \sin(\theta))^2}</math>
<math>E_{net_x} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{1}{(p_x - a)^2 } - \frac{1}{(p_x + a)^2 } \Bigg)</math>


Rearranging, <math>\sqrt{|\vec r|^2 * (\cos(\theta)^2 + \sin(\theta)^2) + \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}} = \sqrt{|\vec r|^2 + \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}} = |r_+|</math>
Finding a common denominator and simplifying:
<math>E_{net_x} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{4p_x a}{(p_x^2 - a^2)^2} \Bigg)</math>


By a similar method, <math>|r_-| = \sqrt{|\vec r|^2 - \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}}</math>
When <math>p_x \gg a</math>, we can assume that <math>p_x^2 - a^2 \approx p_x^2</math>. This approximation gives:
<math>E_{net_x} \approx \frac{1}{4\pi\epsilon_0} \Bigg(\frac{4 a q_+}{p_x^3} \Bigg)</math>


Plugging this into the earlier equation, we get the monstrosity <math>E_{P} = \frac{q_+}{4\pi\epsilon_{0}} \times (\frac{\hat r_{+}}{|\vec r|^2 + \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}} - \frac{\hat r_{-}}{|\vec r|^2 - \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}})</math>
Since the dipole moment is <math>p = qd = 2aq</math>, this simplifies to the famous parallel axis formula:
<math>E_{axis} \approx \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}</math>


To subtract the fractions, we first put them in like terms:
===On the Perpendicular Axis===
On the perpendicular bisector, <math>p_x = 0</math>. The vertical forces from both point charges cancel out, leaving only a horizontal force antiparallel to the dipole moment.


<math> \frac{\hat r_{+}}{|\vec r|^2 + \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}}
Plugging <math>p_x = 0</math> into our exact <math>E_{net_x}</math> formula:
\times
<math> E_{net_x} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{-a}{\Big(a^2 +p_y^2 \Big)^\frac{3}{2} } - \frac{a}{\Big(a^2 +p_y^2 \Big)^\frac{3}{2} } \Bigg) = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{-2a}{\Big(a^2 +p_y^2 \Big)^\frac{3}{2} } \Bigg) </math>
\frac{|\vec r|^2 - \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}}{|\vec r|^2 - \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}})
=
\frac{\hat r_{+}\times (|\vec r|^2 - \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4})}{|\vec r|^4 - \frac{|\vec r|^2 d |\vec r| \cos(\theta)}{2} + \frac{|\vec r|^2 d^2}{4} + \frac{|\vec r|^2 d |\vec r| \cos(\theta)}{2} - \frac{d^2 |\vec r|^2 \cos(\theta)^2}{4} + \frac{d^3|\vec r|cos(\theta)}{8} + \frac{|\vec r|^2 d^2}{4} - \frac{d^3|\vec r|cos(\theta)}{8} + \frac{d^4}{16}}</math>


Cancelling like terms simplifies this to
When distance <math>p_y \gg a</math>, the denominator approximates to <math>(p_y^2)^{3/2} = p_y^3</math>. Substituting <math>p = 2aq</math> yields the perpendicular axis formula:
<math>\frac{\hat r_{+}\times (|\vec r|^2 - \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4})}{|\vec r|^4 + \frac{|\vec r|^2 d^2}{2} -\frac{d^2 |\vec r|^2 \cos(\theta)^2}{4} + \frac{d^4}{16}}</math>
<math>E_{perp} \approx -\frac{1}{4\pi\epsilon_0} \frac{p}{r^3} </math>


==Examples==
==Examples==


Be sure to show all steps in your solution and include diagrams whenever possible
===Simple===
A dipole is located at the origin, composed of charged particles with charge +e and -e, separated by a distance of <math>9 \times 10^{-10}</math> m along the y-axis. The +e charge is on the +y axis. Calculate the net force acting on a single proton located at <math>< 0, 0, 3 \times 10^{-8} ></math> meters.
 
<div class="toccolours mw-collapsible mw-collapsed">
===Click for Solution===
<div class="mw-collapsible-content">
The center of the dipole is at the origin and the observation proton is along the z-axis. Because the observation point is on the perpendicular bisector and <math>r \gg d</math>, we apply the perpendicular approximation formula:
<math>E_{net} = -\frac{1}{4\pi\epsilon_0} \frac{p}{r^3}</math>
 
First, calculate the dipole moment <math>p = qd = (1.6 \times 10^{-19} \text{ C})(9 \times 10^{-10} \text{ m}) = 1.44 \times 10^{-28} \text{ C}\cdot\text{m}</math>.
 
[[File:Phys2212 sample simple.PNG | 300px]]
 
Calculate the electric field:
<math>E = -(9 \times 10^9) \frac{1.44 \times 10^{-28}}{(3 \times 10^{-8})^3} = -48,000 \text{ N/C}</math>
As a vector, this is <math><0, -48000, 0> \text{ N/C}</math>.
 
To find the force on the proton, we use <math>\vec{F} = q\vec{E}</math>:
<math>\vec{F} = (1.6 \times 10^{-19} \text{ C}) \langle0, -48000, 0\rangle = \langle0, -7.68 \times 10^{-15}, 0\rangle \text{ N}</math>.
</div>
</div>


===Simple===
===Middling===
===Middling===
===Difficult===
A ball of mass M and radius R is given an unknown negative charge spread uniformly over its surface. The ball is hanging from a thread and can move freely. A distance L directly below the center of the ball, a small permanent dipole is oriented such that the dipole axis is parallel with the center of the ball. The dipole has a dipole moment <math>p = qs</math>, with a distance s between the positive and negative charges of the dipole, and a mass m. The positive charge of the dipole is oriented closer to the center of the ball.
 
a) Calculate the required charge on the ball to levitate the dipole.
b) If the dipole is turned 90 degrees clockwise without changing its position relative to the ball, what effect does this have on the ball?
 
<div class="toccolours mw-collapsible mw-collapsed">
===Click for Solutions===
<div class="mw-collapsible-content">
'''Part A)''' Because the dipole is small, we assume <math>s \ll L</math>. We want the upward electric force on the dipole to perfectly balance the downward force of gravity (<math>F_G = F_E</math>). By Newton's Third Law, the force exerted on the dipole by the ball is equal and opposite to the force exerted on the ball by the dipole.
 
The electric field from the dipole at the location of the ball (parallel axis) is:
<math>E_{dipole} = \frac{1}{4\pi\epsilon_0} \frac{2p}{L^3}</math>
 
The force on the ball is <math>F_E = |Q|E_{dipole}</math>. Setting this equal to gravity (mg):
<math>mg = |Q| \left( \frac{1}{4\pi\epsilon_0} \frac{2p}{L^3} \right)</math>
 
Solving for <math>|Q|</math>:
<math>|Q| = \frac{mgL^3}{2p} \left(\frac{1}{4\pi\epsilon_0}\right)^{-1}</math>
 
Since the positive charge of the dipole is pointing upward (closer to the ball), the ball must carry a negative charge to create an attractive force capable of lifting the dipole. Therefore:
<math>Q = -\frac{mgL^3}{2p (1 / 4\pi\epsilon_0)}</math>
 
'''Part B)''' Rotating the dipole 90 degrees clockwise shifts the observation location to the perpendicular axis. The positive end of the dipole now points right, and the negative end points left. Consequently, the electric field from the dipole at the ball's location points to the left. Because the ball is negatively charged, it will experience a force in the direction opposite to the electric field, pushing the ball to the '''right'''.
</div>
</div>
 
===Concept Question===
Is it possible for a permanent electric dipole to have a net (total) charge of zero?
 
<div class="toccolours mw-collapsible mw-collapsed">
===Click for Solution===
<div class="mw-collapsible-content">
Yes! By definition, an ideal electric dipole consists of two charges of equal magnitude and opposite signs (+q and -q). Therefore, the net total charge of the system is exactly zero, even though it still produces a highly functional electric field due to the spatial separation of those charges.
</div>
</div>
 
===Practice Test Problem===
[[File:Exampleprac.jpg|400px]]
 
==Electric Field of an Electric Dipole==
The electric field of an electric dipole can be constructed as a vector sum of the point charge fields of the two individual charges. As seen in the graphics below, the electric field lines always point away from the positive particle and towards the negative particle. This is a crucial characteristic used to identify the orientation of an unknown dipole.
 
Direction of electric dipole:
[[File:dipd.gif]]
 
Electric Field:
[[File:edip2.gif]]
 
In introductory physics, most questions evaluate observation locations strictly on the parallel axis or the perpendicular bisector. Decomposing the vectors from each charged particle helps conceptualize this.
 
[[File:Phys2212 dipole electric field.PNG]]
 
==Torque==
===Derivation===
Consider a dipole with an arbitrary orientation in a uniform external electric field:
 
[[File:dipole_torque_2.gif]]
 
The uniform electric field exerts a force on both point charges (<math>\vec{F} = q\vec{E}</math>). Because the field is uniform and the charges are equal and opposite, the net linear force on the dipole is zero. However, because the forces are applied at different points in space, they create a rotational force, or '''torque''', that attempts to align the dipole with the external field.
 
The component of force perpendicular to the dipole axis is <math>F_{\perp}= qE\sin \theta</math>, where <math>\theta</math> is the angle between the electric field and the dipole moment. Generalizing this into a cross product gives the torque on the dipole:
 
<math>\vec{\tau} = \vec{p} \times \vec{E}</math>
Magnitude: <math>|\tau| = pE\sin\theta</math>
 
===Direction===
The direction of the torque vector can be determined using the right-hand rule. It will always be perpendicular to both the dipole axis and the applied electric field. When the dipole aligns perfectly parallel with the electric field (<math>\theta = 0^\circ</math>), the cross product is zero, meaning the dipole experiences zero torque and is in a state of stable equilibrium.
[[File:dipole_t.gif]]
 
===Energy and Work===
The torque that rotates a dipole moves it from a configuration of higher potential energy to lower potential energy.
 
[[File:dipole_torque.gif]]
 
Rotating a dipole against this gradient requires external work. By convention, the potential energy is defined as zero when the dipole is exactly perpendicular to the electric field (<math>\theta = 90^\circ</math>). The potential energy (U) of the system can be calculated using the dot product:
 
<math>U = -pE \cos\theta = -\vec{p} \cdot \vec{E}</math>
 
This indicates that the potential energy is minimized when <math>\cos\theta = 1</math> (the dipole is perfectly parallel to the field) and maximized when <math>\cos\theta = -1</math> (the dipole is perfectly antiparallel).
 
===Nonuniform Electric Field===
If the external electric field is not uniform, the two point charges will experience slightly different forces. This means the net linear force on the dipole will no longer be zero, and the dipole will be physically pulled toward the region where the electric field is strongest.
 
For a non-uniform electric field E with a spatial gradient <math>\nabla</math>, the net force on an ideal dipole can be approximated by the leading term of its power series expansion:
 
<math>\vec{F} = (\vec{p} \cdot \nabla) \vec{E}(r)</math>
 
==Electric Dipole Concept Map==
This concept map illustrates the various fields, forces, and relationships caused by an electric dipole.
 
[[File:dipolecon.gif]]


==Connectedness==
==Connectedness==
#How is this topic connected to something that you are interested in?
Dipoles are foundational across physics, chemistry, and biology. The polarity of molecules is entirely dependent on permanent dipoles. For instance, the dipole moment of water governs its behavior as a universal solvent, directly dictating principles like hydrophilicity and hydrophobicity.
#How is it connected to your major?
#Is there an interesting industrial application?


==History==
In cellular biology, the selectively permeable membrane of human cells relies on this polarity. The outer heads of the lipid bilayer are polar (hydrophilic), while the inner tails are non-polar (hydrophobic). This structure regulates which molecules can enter and exit the cell, maintaining homeostasis.


Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
In Biomedical Engineering, understanding dipole interactions is critical for techniques like ion polarization, cellular separation, and even the design of certain MRI contrast agents.


== See also ==
==Electric Dipoles in Nature==
Beyond classical mechanics, electric dipoles are utilized to probe the fundamental symmetries of the universe. The search for a permanent Electric Dipole Moment (EDM) in fundamental particles like electrons and neutrons is a powerful tool to test for violations in time-reversal (T) and charge-parity (CP) symmetries. Discovering a non-zero EDM in an electron would help physicists explain the cosmic imbalance between matter and antimatter.


Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?
==History==
Electric dipoles have been conceptualized since the mid-1800s. However, atomic dipoles could only be fully understood after Niels Bohr introduced his quantum model of the atom in 1913. This leap in understanding bridged the gap between macro-scale electrostatics and micro-scale atomic chemistry, paving the way for modern solid-state physics.


===Further reading===
== See also ==
 
[http://www.physicsbook.gatech.edu/Magnetic_Dipole Magnetic Dipole]
Books, Articles or other print media on this topic


===External links===
===External links===
 
* [https://en.wikipedia.org/wiki/Electric_charge Electric Charge]
Internet resources on this topic
* [https://en.wikibooks.org/wiki/Physics_Exercises/Electrostatics Additional Dipole Derivations]
* [https://en.wikipedia.org/wiki/Electric_dipole_moment Electric Dipole Moment]
* [https://en.wikipedia.org/wiki/Dipole Dipole]
* [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/diptor.html Electric Dipole Torque]


==References==
==References==
 
* [http://education.jlab.org/qa/historymag_01.html Magnet History]
This section contains the the references you used while writing this page
* [https://en.wikipedia.org/wiki/Bohr_model Bohr Model]
* [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/diph2o.html Electric Dipole]


[[Category:Fields]]
[[Category:Fields]]

Latest revision as of 00:00, 27 April 2026

Claimed By: Yashwin Thammiraju, Spring 2026

Summary

An electric dipole is made up of two point charges that have equal but opposite electric charges (q) and are separated by a short distance (d).

An Electric Dipole

The electric field of a dipole is inversely proportional to the cube of the distance from the dipole ([math]\displaystyle{ 1/r^3 }[/math]), unlike a single point charge which falls off by [math]\displaystyle{ 1/r^2 }[/math]. The field's magnitude and direction are highly dependent on whether you are observing it along the parallel axis (the line separating the two charges) or the perpendicular axis (the bisector).

A temporary dipole can be created when you place a neutral atom in an external electric field. Due to the movement of the electron cloud relative to the nucleus, the atom polarizes (shifting negative charge to one side and positive charge to the other), yielding a separation of charge.

Electric dipoles are characterized by their dipole moment ([math]\displaystyle{ \vec{p} }[/math]), a vector quantity measuring the strength and separation of the positive and negative electrical charges within a system. For two point charges, +q and -q, separated by a distance d, the magnitude of the dipole moment is:

[math]\displaystyle{ p = qd }[/math]

A prime example of a permanent dipole in nature is the water molecule (H2O), which forms a 105-degree angle between the two hydrogen atoms connected to the oxygen. Because oxygen has a greater electronegativity, it pulls more strongly on the shared electrons. Consequently, the oxygen end of the molecule becomes more negatively charged compared to the hydrogen end, and the net electric dipole moment points towards the oxygen atom.

Dipole moment of water

Computational Model

To better visualize the electric field generated by an electric dipole, we can use a computational model. The GlowScript simulation below calculates the exact superposition of the point charges and displays the resulting electric field vectors at various observation locations.

<html><iframe src="https://trinket.io/embed/glowscript/31d0f9ad9e" width="100%" height="400" frameborder="0" marginwidth="0" marginheight="0" allowfullscreen></iframe></html>

Mathematical Models

An Exact Model

Polarization by an electric field

An electric dipole is constructed from two point charges: one at position [math]\displaystyle{ [\frac{d}{2}, 0] }[/math] and one at position [math]\displaystyle{ [\frac{-d}{2}, 0] }[/math]. These point charges have equal and opposite charge magnitudes. We wish to calculate the exact electric field due to the dipole at some observation point P in the plane (see the figure). Point P can be defined by its coordinates [math]\displaystyle{ [p_x, p_y] }[/math] from the midpoint of the dipole, or by a distance r and an angle [math]\displaystyle{ \theta }[/math].

Using the superposition principle, the net electric field at P is [math]\displaystyle{ E_{net} = E_{q_+} + E_{q_-} }[/math]. We can break this down into x and y components:

  • [math]\displaystyle{ E_{net_x} = E_{q_{+x}} + E_{q_{-x}} }[/math]
  • [math]\displaystyle{ E_{net_y} = E_{q_{+y}} + E_{q_{-y}} }[/math]

Let [math]\displaystyle{ \theta_+ }[/math] be the angle from q+ to P. The y-component of the positive charge's field is [math]\displaystyle{ E_{q_{+y}} = E_{q_+} \sin(\theta_+) }[/math].

To find [math]\displaystyle{ \theta_+ }[/math] and its counterpart [math]\displaystyle{ \theta_- }[/math], we look at the geometry. [math]\displaystyle{ \theta_+ }[/math] belongs to a right triangle with an opposite side length of [math]\displaystyle{ p_y }[/math] and an adjacent side length of [math]\displaystyle{ p_x - \frac{d}{2} }[/math]. Therefore: [math]\displaystyle{ \sin(\theta_+) = \frac{p_y}{\sqrt{(p_x - \frac{d}{2})^2+p_y^2}} }[/math]

The denominator here represents the hypotenuse, which is simply the distance [math]\displaystyle{ |\vec r_+| }[/math] from the positive charge to the observation point. A similar geometric analysis for the negative charge gives: [math]\displaystyle{ \sin(\theta_-) = \frac{p_y}{\sqrt{(p_x + \frac{d}{2})^2+p_y^2}} }[/math] where the denominator is [math]\displaystyle{ |\vec r_-| }[/math].

The general formula for the magnitude of an electric field from a point charge is [math]\displaystyle{ |E| = \frac{1}{4\pi\epsilon_0} \frac{q}{|\vec r|^2} }[/math]. Applying this to both charges: [math]\displaystyle{ E_{net_y} = \frac{1}{4\pi\epsilon_0} \frac{q_+}{|\vec r_+|^2} \sin(\theta_+) + \frac{1}{4\pi\epsilon_0} \frac{q_-}{|\vec r_-|^2} \sin(\theta_-) }[/math]

Noting that [math]\displaystyle{ q_+ = -q_- }[/math], we can factor out the charge: [math]\displaystyle{ E_{net_y} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{1}{|\vec r_+|^2}\sin(\theta_+) - \frac{1}{|\vec r_-|^2}\sin(\theta_-)\Bigg) }[/math]

Substituting our expanded radii and sines into the equation yields: [math]\displaystyle{ E_{net_y} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{1}{(p_x - \frac{d}{2})^2 +p_y^2 } \frac{p_y}{\sqrt{(p_x - \frac{d}{2})^2+p_y^2}} - \frac{1}{(p_x + \frac{d}{2})^2 +p_y^2 } \frac{p_y}{\sqrt{(p_x + \frac{d}{2})^2+p_y^2}} \Bigg) }[/math]

Combining the denominators simplifies this to the exact analytical form for the y-component: [math]\displaystyle{ E_{net_y} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{p_y}{\Big((p_x - \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2} } - \frac{p_y}{\Big((p_x + \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2} } \Bigg) }[/math]

The derivation for the x-direction follows the exact same logic, using cosine (adjacent over hypotenuse) instead of sine. The result is: [math]\displaystyle{ E_{net_x} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{p_x - \frac{d}{2}}{\Big((p_x - \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2} } - \frac{p_x + \frac{d}{2}}{\Big((p_x + \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2} } \Bigg) }[/math]

These formulae provide the exact electric field due to an electric dipole anywhere on the 2D plane.

Special Cases (Approximations)

When the observation distance is much greater than the separation distance ([math]\displaystyle{ r \gg d }[/math]), we can simplify the exact models into the standard dipole approximations. Let [math]\displaystyle{ a = \frac{d}{2} }[/math].

On the Parallel Axis

On the axis running through the two charges, [math]\displaystyle{ p_y = 0 }[/math], meaning [math]\displaystyle{ E_{net_y} = 0 }[/math]. Plugging [math]\displaystyle{ p_y = 0 }[/math] into our exact [math]\displaystyle{ E_{net_x} }[/math] formula:

[math]\displaystyle{ E_{net_x} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{1}{(p_x - a)^2 } - \frac{1}{(p_x + a)^2 } \Bigg) }[/math]

Finding a common denominator and simplifying: [math]\displaystyle{ E_{net_x} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{4p_x a}{(p_x^2 - a^2)^2} \Bigg) }[/math]

When [math]\displaystyle{ p_x \gg a }[/math], we can assume that [math]\displaystyle{ p_x^2 - a^2 \approx p_x^2 }[/math]. This approximation gives: [math]\displaystyle{ E_{net_x} \approx \frac{1}{4\pi\epsilon_0} \Bigg(\frac{4 a q_+}{p_x^3} \Bigg) }[/math]

Since the dipole moment is [math]\displaystyle{ p = qd = 2aq }[/math], this simplifies to the famous parallel axis formula: [math]\displaystyle{ E_{axis} \approx \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} }[/math]

On the Perpendicular Axis

On the perpendicular bisector, [math]\displaystyle{ p_x = 0 }[/math]. The vertical forces from both point charges cancel out, leaving only a horizontal force antiparallel to the dipole moment.

Plugging [math]\displaystyle{ p_x = 0 }[/math] into our exact [math]\displaystyle{ E_{net_x} }[/math] formula: [math]\displaystyle{ E_{net_x} = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{-a}{\Big(a^2 +p_y^2 \Big)^\frac{3}{2} } - \frac{a}{\Big(a^2 +p_y^2 \Big)^\frac{3}{2} } \Bigg) = \frac{q_+}{4\pi\epsilon_0} \Bigg(\frac{-2a}{\Big(a^2 +p_y^2 \Big)^\frac{3}{2} } \Bigg) }[/math]

When distance [math]\displaystyle{ p_y \gg a }[/math], the denominator approximates to [math]\displaystyle{ (p_y^2)^{3/2} = p_y^3 }[/math]. Substituting [math]\displaystyle{ p = 2aq }[/math] yields the perpendicular axis formula: [math]\displaystyle{ E_{perp} \approx -\frac{1}{4\pi\epsilon_0} \frac{p}{r^3} }[/math]

Examples

Simple

A dipole is located at the origin, composed of charged particles with charge +e and -e, separated by a distance of [math]\displaystyle{ 9 \times 10^{-10} }[/math] m along the y-axis. The +e charge is on the +y axis. Calculate the net force acting on a single proton located at [math]\displaystyle{ \lt 0, 0, 3 \times 10^{-8} \gt }[/math] meters.

Click for Solution

The center of the dipole is at the origin and the observation proton is along the z-axis. Because the observation point is on the perpendicular bisector and [math]\displaystyle{ r \gg d }[/math], we apply the perpendicular approximation formula: [math]\displaystyle{ E_{net} = -\frac{1}{4\pi\epsilon_0} \frac{p}{r^3} }[/math]

First, calculate the dipole moment [math]\displaystyle{ p = qd = (1.6 \times 10^{-19} \text{ C})(9 \times 10^{-10} \text{ m}) = 1.44 \times 10^{-28} \text{ C}\cdot\text{m} }[/math].

Calculate the electric field: [math]\displaystyle{ E = -(9 \times 10^9) \frac{1.44 \times 10^{-28}}{(3 \times 10^{-8})^3} = -48,000 \text{ N/C} }[/math] As a vector, this is [math]\displaystyle{ \lt 0, -48000, 0\gt \text{ N/C} }[/math].

To find the force on the proton, we use [math]\displaystyle{ \vec{F} = q\vec{E} }[/math]: [math]\displaystyle{ \vec{F} = (1.6 \times 10^{-19} \text{ C}) \langle0, -48000, 0\rangle = \langle0, -7.68 \times 10^{-15}, 0\rangle \text{ N} }[/math].

Middling

A ball of mass M and radius R is given an unknown negative charge spread uniformly over its surface. The ball is hanging from a thread and can move freely. A distance L directly below the center of the ball, a small permanent dipole is oriented such that the dipole axis is parallel with the center of the ball. The dipole has a dipole moment [math]\displaystyle{ p = qs }[/math], with a distance s between the positive and negative charges of the dipole, and a mass m. The positive charge of the dipole is oriented closer to the center of the ball.

a) Calculate the required charge on the ball to levitate the dipole. b) If the dipole is turned 90 degrees clockwise without changing its position relative to the ball, what effect does this have on the ball?

Click for Solutions

Part A) Because the dipole is small, we assume [math]\displaystyle{ s \ll L }[/math]. We want the upward electric force on the dipole to perfectly balance the downward force of gravity ([math]\displaystyle{ F_G = F_E }[/math]). By Newton's Third Law, the force exerted on the dipole by the ball is equal and opposite to the force exerted on the ball by the dipole.

The electric field from the dipole at the location of the ball (parallel axis) is: [math]\displaystyle{ E_{dipole} = \frac{1}{4\pi\epsilon_0} \frac{2p}{L^3} }[/math]

The force on the ball is [math]\displaystyle{ F_E = |Q|E_{dipole} }[/math]. Setting this equal to gravity (mg): [math]\displaystyle{ mg = |Q| \left( \frac{1}{4\pi\epsilon_0} \frac{2p}{L^3} \right) }[/math]

Solving for [math]\displaystyle{ |Q| }[/math]: [math]\displaystyle{ |Q| = \frac{mgL^3}{2p} \left(\frac{1}{4\pi\epsilon_0}\right)^{-1} }[/math]

Since the positive charge of the dipole is pointing upward (closer to the ball), the ball must carry a negative charge to create an attractive force capable of lifting the dipole. Therefore: [math]\displaystyle{ Q = -\frac{mgL^3}{2p (1 / 4\pi\epsilon_0)} }[/math]

Part B) Rotating the dipole 90 degrees clockwise shifts the observation location to the perpendicular axis. The positive end of the dipole now points right, and the negative end points left. Consequently, the electric field from the dipole at the ball's location points to the left. Because the ball is negatively charged, it will experience a force in the direction opposite to the electric field, pushing the ball to the right.

Concept Question

Is it possible for a permanent electric dipole to have a net (total) charge of zero?

Click for Solution

Yes! By definition, an ideal electric dipole consists of two charges of equal magnitude and opposite signs (+q and -q). Therefore, the net total charge of the system is exactly zero, even though it still produces a highly functional electric field due to the spatial separation of those charges.

Practice Test Problem

Electric Field of an Electric Dipole

The electric field of an electric dipole can be constructed as a vector sum of the point charge fields of the two individual charges. As seen in the graphics below, the electric field lines always point away from the positive particle and towards the negative particle. This is a crucial characteristic used to identify the orientation of an unknown dipole.

Direction of electric dipole:

Electric Field:

In introductory physics, most questions evaluate observation locations strictly on the parallel axis or the perpendicular bisector. Decomposing the vectors from each charged particle helps conceptualize this.

Torque

Derivation

Consider a dipole with an arbitrary orientation in a uniform external electric field:

The uniform electric field exerts a force on both point charges ([math]\displaystyle{ \vec{F} = q\vec{E} }[/math]). Because the field is uniform and the charges are equal and opposite, the net linear force on the dipole is zero. However, because the forces are applied at different points in space, they create a rotational force, or torque, that attempts to align the dipole with the external field.

The component of force perpendicular to the dipole axis is [math]\displaystyle{ F_{\perp}= qE\sin \theta }[/math], where [math]\displaystyle{ \theta }[/math] is the angle between the electric field and the dipole moment. Generalizing this into a cross product gives the torque on the dipole:

[math]\displaystyle{ \vec{\tau} = \vec{p} \times \vec{E} }[/math] Magnitude: [math]\displaystyle{ |\tau| = pE\sin\theta }[/math]

Direction

The direction of the torque vector can be determined using the right-hand rule. It will always be perpendicular to both the dipole axis and the applied electric field. When the dipole aligns perfectly parallel with the electric field ([math]\displaystyle{ \theta = 0^\circ }[/math]), the cross product is zero, meaning the dipole experiences zero torque and is in a state of stable equilibrium.

Energy and Work

The torque that rotates a dipole moves it from a configuration of higher potential energy to lower potential energy.

Rotating a dipole against this gradient requires external work. By convention, the potential energy is defined as zero when the dipole is exactly perpendicular to the electric field ([math]\displaystyle{ \theta = 90^\circ }[/math]). The potential energy (U) of the system can be calculated using the dot product:

[math]\displaystyle{ U = -pE \cos\theta = -\vec{p} \cdot \vec{E} }[/math]

This indicates that the potential energy is minimized when [math]\displaystyle{ \cos\theta = 1 }[/math] (the dipole is perfectly parallel to the field) and maximized when [math]\displaystyle{ \cos\theta = -1 }[/math] (the dipole is perfectly antiparallel).

Nonuniform Electric Field

If the external electric field is not uniform, the two point charges will experience slightly different forces. This means the net linear force on the dipole will no longer be zero, and the dipole will be physically pulled toward the region where the electric field is strongest.

For a non-uniform electric field E with a spatial gradient [math]\displaystyle{ \nabla }[/math], the net force on an ideal dipole can be approximated by the leading term of its power series expansion:

[math]\displaystyle{ \vec{F} = (\vec{p} \cdot \nabla) \vec{E}(r) }[/math]

Electric Dipole Concept Map

This concept map illustrates the various fields, forces, and relationships caused by an electric dipole.

Connectedness

Dipoles are foundational across physics, chemistry, and biology. The polarity of molecules is entirely dependent on permanent dipoles. For instance, the dipole moment of water governs its behavior as a universal solvent, directly dictating principles like hydrophilicity and hydrophobicity.

In cellular biology, the selectively permeable membrane of human cells relies on this polarity. The outer heads of the lipid bilayer are polar (hydrophilic), while the inner tails are non-polar (hydrophobic). This structure regulates which molecules can enter and exit the cell, maintaining homeostasis.

In Biomedical Engineering, understanding dipole interactions is critical for techniques like ion polarization, cellular separation, and even the design of certain MRI contrast agents.

Electric Dipoles in Nature

Beyond classical mechanics, electric dipoles are utilized to probe the fundamental symmetries of the universe. The search for a permanent Electric Dipole Moment (EDM) in fundamental particles like electrons and neutrons is a powerful tool to test for violations in time-reversal (T) and charge-parity (CP) symmetries. Discovering a non-zero EDM in an electron would help physicists explain the cosmic imbalance between matter and antimatter.

History

Electric dipoles have been conceptualized since the mid-1800s. However, atomic dipoles could only be fully understood after Niels Bohr introduced his quantum model of the atom in 1913. This leap in understanding bridged the gap between macro-scale electrostatics and micro-scale atomic chemistry, paving the way for modern solid-state physics.

See also

Magnetic Dipole

External links

References

Retrieved from "http://www.physicsbook.gatech.edu/index.php?title=Electric_Dipole&oldid=48165"