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An '''Electric Dipole''' is a pair of equal and opposite [[Point Charge]]s separated by a small distance. Electric dipoles have a number of interesting properties.
'''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>.


===An Exact Model===
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:
[[File:Dipole.png|300px|thumb|An Electric Dipole]]
* <math>E_{net_x} = E_{q_{+x}} + E_{q_{-x}}</math>
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 are of equal and opposite charge. We then wish to know the electric field due to the dipole at some point <math>p</math> in the plane (see the figure). <math>p</math> can be considered either a distance <math>[x_0, y_0]</math> from the midpoint of the dipole, or a distance <math>r</math> and an angle <math>\theta</math> as in the diagram.
* <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>.
 
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:
<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>
 
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>
 
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>
 
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>
 
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>
 
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>r \gg d</math>), we can simplify the exact models into the standard dipole approximations. Let <math>a = \frac{d}{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:
 
<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>


We state that the net electric field at <math>p</math> is <math>E_{net}</math> and has an x and y component, <math>E_{net_x}</math> and <math>E_{net_y}</math>. Then we can individually calculate the x and y components. First we realize that since <math>E_{net} = E_{q_+} + E_{q_-}</math>, <math>E_{net_x} = E_{q_{+x}} + E_{q_{-x}}</math>, similarly for y <math>E_{net_y} = E_{q_{+y}} + E_{q_{-y}}</math>. At this point, its worth noting that <math>E_{q_{+y}} = E_{q_+} * cos(\theta_+)</math>, where <math>\theta_+</math> is the angle from <math>q_{+}</math> to <math>p</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>


<math>\theta_+</math> and its counterpart <math>\theta_-</math> are not known. However, we can calculate them. We know <math>\theta_+</math> is formed by a triangle with one side length <math>p_y</math> and one side length <math>p_x - \frac{d}{2}</math>. Then <math>sin(\theta_+) = \frac{p_y}{\sqrt{(p_x - \frac{d}{2})^2+p_y^2}}</math>, from which you can calculate the angle. This looks disgusting, but a close inspection shows that <math>p_y</math> is the opposite side of the triangle, and the denominator is an expression forming the hypotenuse of the triangle (<math>r_+</math>) from known quantities. A similar method shows that <math>sin(\theta_-) = \frac{p_y}{\sqrt{(p_x + \frac{d}{2})^2+p_y^2}}</math>, where once again <math>\sqrt{(p_x + \frac{d}{2})^2+p_y^2} = |\vec r_-|</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>


We now have values for <math> d, q, \theta_+, \theta_-, \vec r_+, \vec r_-</math>. This is enough to calculate <math>E_{net}</math> in both directions. The general formula for electric field strength from a [[Point Charge]] is <math>E = \frac{1}{4\pi\epsilon_0} \frac{q}{|\vec r|^2} \hat r</math>. Then <math>|E_+| = \frac{1}{4\pi\epsilon_0} \frac{q_+}{|\vec r_+|^2}</math> and <math>|E_-| = \frac{1}{4\pi\epsilon_0} \frac{q_-}{|\vec r_-|^2}</math>. We want solely the magnitude in this case because we can calculate direction and component forces using sin and cosine. Its worth noting that we can expand <math>r_+, r_-</math> to the form in the denominator of the sine and cosine. We will use this later.
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>


First we calculate <math>E_{net_y}</math>. <math>E_{net_y} = E_{+_y} + E_{-_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>.
===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.  


Then we combine some terms, noting that <math> q_+ = -q_-</math>. <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>
Plugging <math>p_x = 0</math> into our exact <math>E_{net_x}</math> formula:
<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>


Now it gets ugly, we expand our radii and sines. To recap, <math>sin(\theta_+) = \frac{p_y}{\sqrt{(p_x - \frac{d}{2})^2+p_y^2}}</math>, <math>sin(\theta_-) = \frac{p_y}{\sqrt{(p_x + \frac{d}{2})^2+p_y^2}}</math>, <math>|r_+| = \sqrt{(p_x - \frac{d}{2})^2 +p_y^2}</math> and <math>|r_-| = \sqrt{(p_x + \frac{d}{2})^2 +p_y^2}</math>, giving us
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>E_{perp} \approx -\frac{1}{4\pi\epsilon_0} \frac{p}{r^3} </math>


<math>E_{net_y} =  
==Examples==
\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>


Finally we can combine more terms, the denominators of the expanded sines are the square roots of the radii. We can also pull out the negative sign.
===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.


<math>E_{net_y} =  
<div class="toccolours mw-collapsible mw-collapsed">
\frac{q_+}{4\pi\epsilon_0}
===Click for Solution===
\Bigg(
<div class="mw-collapsible-content">
    \frac{p_y}{
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:
        \Big((p_x - \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2}
<math>E_{net} = -\frac{1}{4\pi\epsilon_0} \frac{p}{r^3}</math>
    }
-
    \frac{p_y}{
        \Big((p_x + \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2}
    }
       
\Bigg)</math> That's as simplified as possible.


Much of the derivation for the x direction is similar. The major difference is that instead of calculating the sine, opposite over hypotenuse, we want cosine, adjacent over hypotenuse. That is, where <math>sin(\theta_+) = \frac{p_y}{\sqrt{(p_x - \frac{d}{2})^2+p_y^2}}</math>, <math>cos(\theta_+) = \frac{p_x - \frac{d}{2}}{\sqrt{(p_x - \frac{d}{2})^2+p_y^2}}</math>. By using this and its counterpart for <math>\theta_-</math>, the result is that
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>.


<math>E_{net_x} =
[[File:Phys2212 sample simple.PNG | 300px]]
\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 provide exact formulae for the electric field due to an electric dipole anywhere on the two-dimensional plane, and they translate easily into 3-dimensions.


==Special Cases==
Calculate the electric field:
We can simplify the solution for many cases
<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>


===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>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.


===On the Parallel Axis===
a) Calculate the required charge on the ball to levitate the dipole.
On the parallel axis, we begin with the now known formula <math>E_{net_x} =  
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?
\frac{q_+}{4\pi\epsilon_0}
 
\Bigg(
<div class="toccolours mw-collapsible mw-collapsed">
    \frac{p_x - \frac{d}{2}}{
===Click for Solutions===
        \Big((p_x - \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2}
<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.
-
 
    \frac{p_x + \frac{d}{2}}{
The electric field from the dipole at the location of the ball (parallel axis) is:
        \Big((p_x + \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2}
<math>E_{dipole} = \frac{1}{4\pi\epsilon_0} \frac{2p}{L^3}</math>
    }
       
\Bigg)</math>. Since we are on the parallel axis, we know that <math>E_{net_y} = 0</math>, and <math>p_y = 0</math>.


Simplifies to  
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>


<math>E_{net_x} =  
Solving for <math>|Q|</math>:
\frac{q_+}{4\pi\epsilon_0}
<math>|Q| = \frac{mgL^3}{2p} \left(\frac{1}{4\pi\epsilon_0}\right)^{-1}</math>
\Bigg(
    \frac{p_x - \frac{d}{2}}{
        \Big((p_x - \frac{d}{2})^2 \Big)^\frac{3}{2}
    }
-  
    \frac{p_x + \frac{d}{2}}{
        \Big((p_x + \frac{d}{2})^2 \Big)^\frac{3}{2}
    }
       
\Bigg)</math>.


Then, combining exponents and reducing the fraction:  
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>E_{net_x} =  
<math>Q = -\frac{mgL^3}{2p (1 / 4\pi\epsilon_0)}</math>
\frac{q_+}{4\pi\epsilon_0} 
\Bigg(
    \frac{1}{
(p_x - \frac{d}{2})^2
    }
-
    \frac{1}{
(p_x + \frac{d}{2})^2
    }
       
\Bigg)</math>.


Then, we can combine these fractions. to simplify the calculations, replace <math>\frac{d}{2}</math> with <math>a</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>


<math>E_{net_x} =  
===Concept Question===
\frac{q_+}{4\pi\epsilon_0} 
Is it possible for a permanent electric dipole to have a net (total) charge of zero?
\Bigg(
    \frac{1}{
(p_x - a)^2
    }
-
    \frac{1}{
(p_x + a)^2
    }
       
\Bigg) =


\frac{q_+}{4\pi\epsilon_0} 
<div class="toccolours mw-collapsible mw-collapsed">
\Bigg(\frac{4p_x a}{(p_x^2 + a^2)^2}
===Click for Solution===
\Bigg)
<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]]


\frac{q_+ 4 a}{4\pi\epsilon_0} 
==Electric Field of an Electric Dipole==
\Bigg(\frac{p_x}{(p_x^2 + a^2)^2}
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.
\Bigg)
</math>.


This is the formula. When <math>p_x >> a</math>, we can assume that <math>p_x^2 + a^2</math> is very close to <math>p_x^2</math>. Then
Direction of electric dipole:
[[File:dipd.gif]]


<math>E_{net_x} \approx
Electric Field:
\frac{q_+ 4 a}{4\pi\epsilon_0} 
[[File:edip2.gif]]
\Bigg(\frac{p_x}{(p_x^2)^2}
\Bigg) =


\frac{q_+ 4 a}{4\pi\epsilon_0} 
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.
\Bigg(\frac{p_x}{p_x^4}
\Bigg)
=
\frac{1}{4\pi\epsilon_0} 
\Bigg(\frac{4 a q_+}{p_x^3}
\Bigg)</math>


===On the Perpendicular Axis===
[[File:Phys2212 dipole electric field.PNG]]
We can do a similar simplification for the perpendicular axis.  We know that <math>E_{net_y} = 0</math> because the vertical forces from both point charges cancel, leaving only horizontal forces.


<math>
==Torque==
E_{net_x} =  
===Derivation===
\frac{q_+}{4\pi\epsilon_0} 
Consider a dipole with an arbitrary orientation in a uniform external electric field:
\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>


In this case though, <math>p_x = 0</math>
[[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.


<math>
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:
E_{net_x} =  
\frac{q_+}{4\pi\epsilon_0} 
\Bigg(
    \frac{- \frac{d}{2}}{
        \Big(( - \frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2}
    }
-
    \frac{\frac{d}{2}}{
        \Big((\frac{d}{2})^2 +p_y^2 \Big)^\frac{3}{2}
    }
       
\Bigg)</math>


<math>\vec{\tau} = \vec{p} \times \vec{E}</math>
Magnitude: <math>|\tau| = pE\sin\theta</math>


Once again, we say <math>a = \frac{d}{2}</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]]


<math>
===Energy and Work===
E_{net_x} =  
The torque that rotates a dipole moves it from a configuration of higher potential energy to lower potential energy.
\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{-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>


And this is our result.
[[File:dipole_torque.gif]]


Once again, when <math>d</math> is much smaller than <math> p_y</math>, <math>a</math> is also small, so we can assume that the denominator is just <math>p_y</math>. This allows us to simplify the resulting equation to
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>E_{net_x} = \frac{q_+}{4\pi\epsilon_0} \frac{-2a}{p_y^3} </math>
<math>U = -pE \cos\theta = -\vec{p} \cdot \vec{E}</math>


==Examples==
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).


Be sure to show all steps in your solution and include diagrams whenever possible
===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.


===Simple===
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:
===Middling===
===Difficult===


==Connectedness==
<math>\vec{F} = (\vec{p} \cdot \nabla) \vec{E}(r)</math>
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?


==History==
==Electric Dipole Concept Map==
This concept map illustrates the various fields, forces, and relationships caused by an electric dipole.


Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
[[File:dipolecon.gif]]


== See also ==
==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.


===External links===
==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.


[https://en.wikibooks.org/wiki/Physics_Exercises/Electrostatics Additional Dipole Derivations]
== See also ==
[http://www.physicsbook.gatech.edu/Magnetic_Dipole Magnetic Dipole]


[https://en.wikipedia.org/wiki/Electric_dipole_moment Electric Dipole Moment]
===External links===
* [https://en.wikipedia.org/wiki/Electric_charge Electric Charge]
* [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

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