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Claimed by jmorton32 (2015) and edited by Shivani (Spring 2016)
'''Claimed By: Yashwin Thammiraju, Spring 2026'''
and claimed by Hyder Hasnain (Fall 2016) 
 
Claimed by Joshua Jacob (Fall 2017)
Edited Further by Shivani Mehrotra (Fall 2017)
==Summary==
==Summary==


An electric dipole is made up of two point charges that have equal but opposite electric charges and are separated by a distance.
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|300px|thumb|An Electric Dipole]]
[[File:dipo.jpg|An Electric Dipole]]


The electric field is proportional to the cube of the distance from the dipole, and is dependent on whether you’re moving along the line separating the two charges or perpendicular to it.  
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 dipole can be created, for example, when you place a neutral atom in an electric field, because the positively-charged parts of the atom will be pulled one way, and the negatively-charged parts the other way, creating a separation of charge in the direction of the field. However, electric dipoles are not limited to just atoms.  Certain molecules in nature also experience the effects of electric dipoles.  A prime example of this is the molecule for water, which forms a 105 degree angle between the two hydrogens connected to the oxygen.  Since the oxygen has a greater electronegativity, it pulls more strongly on the electrons shared by the oxygen and hydrogen atoms and that end of the molecule becomes more negatively charged compared to the hydrogen end. Therefore, the net electric dipole points towards the oxygen atom.  Therefore, there are two Electric dipoles are particularly useful in atoms and molecules where the effects of charge separation are measurable, but the distance between the particles is too small to quantify.  An electric dipole faces a force of zero in a constant electric field.  However, when a dipole moment is not aligned with the electric field, the dipole is acted on by a torque, which causes rotational movement.
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.  


==Mathematical Models==
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>


===An Exact Model===
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:Dipole.png|300px|thumb|An Electric Dipole]]
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.


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>.
[[File:Water.png|300px|thumb|Dipole moment of water]]


<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>.
==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.  


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


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>.
==Mathematical Models==


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>
===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>.


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
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>


<math>E_{net_y} =
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>.
\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.
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>


<math>E_{net_y} =
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:
\frac{q_+}{4\pi\epsilon_0} 
<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>.
\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> 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
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>


<math>E_{net_x} =  
Noting that <math> q_+ = -q_-</math>, we can factor out the charge:
\frac{q_+}{4\pi\epsilon_0}
<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>
\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==
Substituting our expanded radii and sines into the equation yields:
We can simplify the solution for many cases
<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>


===On the Parallel Axis===
These formulae provide the exact electric field due to an electric dipole anywhere on the 2D plane.
On the parallel axis, we begin with the now known formula <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>. Since we are on the parallel axis, we know that <math>E_{net_y} = 0</math>, and <math>p_y = 0</math>.


Simplifies to
==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>E_{net_x} =  
===On the Parallel Axis===
\frac{q_+}{4\pi\epsilon_0} 
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:
\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:
<math>E_{net_x} =  
\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>.
 
<math>E_{net_x} =
\frac{q_+}{4\pi\epsilon_0} 
\Bigg(
    \frac{1}{
(p_x - a)^2
    }
-
    \frac{1}{
(p_x + a)^2
    }
       
\Bigg) =


\frac{q_+}{4\pi\epsilon_0}
<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>
\Bigg(\frac{4p_x a}{(p_x^2 + a^2)^2}
\Bigg)


=  
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>


\frac{q_+ 4 a}{4\pi\epsilon_0}
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:
\Bigg(\frac{p_x}{(p_x^2 + a^2)^2}
<math>E_{net_x} \approx \frac{1}{4\pi\epsilon_0} \Bigg(\frac{4 a q_+}{p_x^3} \Bigg)</math>
\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
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>
<math>E_{net_x} \approx
\frac{q_+ 4 a}{4\pi\epsilon_0} 
\Bigg(\frac{p_x}{(p_x^2)^2}
\Bigg) =
 
\frac{q_+ 4 a}{4\pi\epsilon_0} 
\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===
===On the Perpendicular Axis===
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.
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>
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>
 
In this case though, <math>p_x = 0</math>
 
 
<math>
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>
 


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


<math>
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:
E_{net_x} =
<math>E_{perp} \approx -\frac{1}{4\pi\epsilon_0} \frac{p}{r^3} </math>
\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.
 
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
 
<math>E_{net_x} \approx \frac{q_+}{4\pi\epsilon_0} \frac{-2a}{p_y^3} </math>


==Examples==
==Examples==


===Simple===
===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.
A dipole is located at the origin, and is composed of charged particles with charge <math>+e</math> and <math>-e</math>, separated by a distance <math>9 \times10^{-10}</math> along the <math>y</math> axis. The <math>+e</math> charge is on the <math>+y</math> axis. Calculate the force on a proton due to this dipole at a location <math>< 0, 0, 3 \times 10^{-8} ></math> meters.


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===Click for Solution===
===Click for Solution===
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The center of the dipole is at the origin and there is a proton along the z axis. In this case, we apply the perpendicular from of the electric field equation. In this case, since <math>r >> d</math>, we can also use an approximate solution. Therefore, we apply the formula <math>E_{net} = \frac{q}{4\pi\epsilon_0} \frac{-2a}{r^3}</math>. Since <math>a = \frac{d}{2}</math>, and r is the distance to the proton, we can plug in the values and solve for the net electric field.
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]]


<math>1.6\times 10^{-19} \times 9 \times 10^9  
Calculate the electric field:
\frac{-9 \times 10^{-10}}
<math>E = -(9 \times 10^9) \frac{1.44 \times 10^{-28}}{(3 \times 10^{-8})^3} = -48,000 \text{ N/C}</math>
{3 \times 10^{-8^3}} = -48000 \frac{N}{C}</math> on the y axis, as a vector: <math><0, -48000, 0></math>.
As a vector, this is <math><0, -48000, 0> \text{ N/C}</math>.


However, we aren't done since we want to know the force. We know that <math>F = qE</math> and in this case, both <math>q</math>, the charge on the proton and <math>E</math>, the electric field, are known. Thus the solution is <math>-48000 \times 1.6 \times 10^{-19} = -7.68 \times 10^{-15}</math> on the y axis, or <math><0, -7.68 \times 10^{-15}, 0></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>
</div>
</div>


===Middling===
===Middling===
A ball of mass <math>M</math> and radius <math>R</math> is given an unknown negative charge spread uniformly over its surface. The ball is hanging from a thread and can move freely. A distance <math>L</math> 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 <math>s</math> between the positive and negative charges of the dipole, and a mass <math>m</math>. The positive charge of the dipole is oriented closer to the center of the ball.
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 charge on the ping-pongball to levitate the dipole
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?
b) the dipole is turned 90 degrees clockwise, without changing its position relative to the ball, what effect does this have on the ball?


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a) Because the dipole is small, we can assume that <math> s << L </math>. We wish to find the force on the dipole such that it can equal the force due to gravity. Once again, <math>F = qE</math> since by newton's third law, for a force exerted on the ball by the dipole, there is an equal and opposite for exerted on the dipole by the ball. That is <math>F_G = F_E</math>, so <math>qE = mg</math> (where <math>g</math> is the acceleration due to gravity). Therefore, in this case we wish to find the force on the ball, meaning the electric field from the dipole and the charge on the ball, <math>Q</math>. The field from the dipole is, since we are on the parallel axis, <math>E = \frac{1}{4\pi\epsilon_0} \frac{2p}{L^3}</math>. Putting this together, we get <math>mg = |Q| \frac{1}{4\pi\epsilon_0} \frac{2p}{L^3}</math>
'''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| = \Bigg(\frac{1}{4\pi\epsilon_0}\Bigg)^{-1} \frac{mgL^3}{2p}</math>.
Solving for <math>|Q|</math>:
<math>|Q| = \frac{mgL^3}{2p} \left(\frac{1}{4\pi\epsilon_0}\right)^{-1}</math>


However, we know that since the positive charge of the dipole is closer to the ball, the charge on the ball must be negative to create an attractive force. <math>|Q| > 0</math>, so our final answer is <math>Q = -\Bigg(\frac{1}{4\pi\epsilon_0}\Bigg)^{-1} \frac{mgL^3}{2p}</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>


b) By rotating the dipole clockwise the direction of the electric field at the location of the ping-pong ball changes. Since the positive end of the dipole is to the right, and the negative end to the left of the dipole, the electric field from the dipole acting on the ball is oriented toward the left. However, since the ball has negative charge, this results in a force on the ball to the right.
'''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'''.
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Permanent dipoles occur when two atoms in a molecule have a great difference in their electronegativity; one atom attracts electrons more than the other, becoming more negative, while the other atom becomes more positive. A permanent magnet, such as a bar magnet, owes its magnetism to the magnetic dipole moment of the electron. A molecule with a permanent dipole moment is called a polar molecule. A molecule is polarized when it carries an induced dipole. A non-degenerate (S-state) atom can have only a zero permanent dipole.  
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.
 
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===Practice Test Problem===
[[File:Exampleprac.jpg|400px]]


==Electric Field of an Electric Dipole==
==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 charges. As can be seen in the graphics, the electric field always points towards the negative particle and points away from the positive particle. This is an important characteristic which can be used to determine which end is positive and which is negative in a 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:
Direction of electric dipole:
[[File:dipd.gif]]
[[File:dipd.gif]]


Electric Field:
Electric Field:
[[File:edip2.gif]]


[[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==
==Torque==
===Derivation===
===Derivation===
Consider a dipole with the following arbitrary orientation in a uniform electric field:
Consider a dipole with an arbitrary orientation in a uniform external electric field:


[[File:dipole_torque_2.gif]]
[[File:dipole_torque_2.gif]]


Note how the electric field will exert an an electric force on each of the point charges. Since the electric field is uniform and charge of each point charge is equal and opposite, the electric force exerted on each point charge will be equal and opposite. The magnitude of each force is simply the force on a point charge, or <math> F = qE </math>. The component of this force perpendicular to the dipole axis can be written as <math> F_{\bot}= qE\sin \theta</math>, where theta is the angle between the electric field and the dipole. It is this perpendicular force which causes rotational motion, and thus is the force component of the applied torque. Since the forces are separated by the distance of the dipole, it can be generalized that an electric field produces the following torque on an electric dipole:
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> \tau\ = p \times E = Eqd\sin\theta</math>  
<math>\vec{\tau} = \vec{p} \times \vec{E}</math>
Magnitude: <math>|\tau| = pE\sin\theta</math>


===Direction===
===Direction===
The direction of the torque can be found using the right hand rule, as it will always be perpendicular to the dipole axis and applied force. The cross product relationship also indicates that when the dipole is parallel to the electric field, no torque will be acting on it.  
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]]
[[File:dipole_t.gif]]
It is important to note that since the forces on each of the point charges are equal and opposite, the net force on the dipole in a uniform field is 0, which also explains why there is no linear motion by the dipole. Also, the torque from the electric field will align the orientation of the dipole to be parallel with the electric field. This is because at this point the force applied to each of the point charges is 0 because <math> F = Eqd \sin 0 = 0 </math>.


===Energy and Work===
===Energy and Work===
 
The torque that rotates a dipole moves it from a configuration of higher potential energy to lower potential energy.
The torque that rotates the dipole moves the orientation of the dipole from a higher energy configuration to a lower energy configuration as shown below.


[[File:dipole_torque.gif]]
[[File:dipole_torque.gif]]


This also means that any rotation done against this energy gradient requires work. The amount of work required to rotate a dipole from its low energy configuration to an angle <math>\theta_0</math> can be derived as:
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>\int_{0}^{\theta_0}\tau d\theta = \int_{0}^{\theta_0}Eqd\sin\theta d\theta  = \int_{0}^{\theta_0}Ep\sin\theta d\theta = Ep(1-\cos\theta_0)</math>
 
Calculating potential energy is a matter of convention. The standard convention it the potential energy is zero when the dipole is perpendicular to the electric field. With this in mind, the potential energy of a dipole can be calculated as:


<math> U = Ep \cos\theta = -p \dot\ E </math>
<math>U = -pE \cos\theta = -\vec{p} \cdot \vec{E}</math>


Again, this would indicate the potential energy is maximized when <math> \cos\theta = 1 </math>, or when the dipole is in the low energy configuration.
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===
===Nonuniform Electric Field===
When the electric field is not uniform, the two point charges will not feel equal and opposite forces at all times, meaning that the net force on the dipole will not be zero. The force on the dipole will be in the direction of where the electric field has the steepest increase. The derivation for the net force on a dipole in a non-uniform electric field is extremely complex [http://bolvan.ph.utexas.edu/~vadim/classes/17f/dipole.pdf], but can be explained in simple terms. Consider an non-uniform electric field E with a gradient <math>\nabla</math>. For any dipole placed in this system, the difference between the electric field at small intervals between the two point charges can be expanded as a power series to calculate the total difference in electric field. It can be proven that in an ideal dipole, all the subleading terms vanish, and all that is left is the leading term of this series. Thus, the general formula for net force on a dipole in a non-uniform electric field can be written as:
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.  


<math> F = ( p \dot\ \nabla ) E(r) </math>
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:


where <math> E(r) </math> is the electric field at the center of the dipole.
<math>\vec{F} = (\vec{p} \cdot \nabla) \vec{E}(r)</math>


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


[[File:dipolecon.gif]]
[[File:dipolecon.gif]]


==Connectedness==
==Connectedness==
Dipoles are incredibly common in physics, chemistry, and other natural sciences. While not specific to electric dipoles, much of the mathematics taught in advanced algorithms is relevant to the study of dipoles in nature, specifically certain randomized algorithms useful in computer science can be used to effectively simulate and predict natural phenomena having to do with dipole forces and the arrangement of many dipoles. Dipoles are useful in determining the behavior of certain molecules with each other.  Polar molecules can act as electric dipoles, such as the water molecule mentioned earlier.  This gives polar molecules certain properties when in a solution.  Dipoles are the basis for polarity in molecules, which leads to other important properties such as hydrophilicity, which is very important in industry as well as in your body.  The cells in the body are surrounded by a selectively permeable membrane. The outer and inner ends of this membrane are polar, while the middle part is non polar. This polarity is very important in determining which molecules enter and exit the cells in our body and therefore how the cells maintain homeostasis. Dipoles are also very common in regards to magnets, which have several applications including Maglev trains and even metal detectors.
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==
==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.
Electric dipoles have been understood since the mid to late 1800s. However, atomic dipoles could only be understood after the discovery of the correct model of the atom by Bohr in 1913. Based on this knowledge, atomic dipoles were used in a lot of technology.  Even though electric dipoles are a newer concept, the human understanding of magnetic dipoles goes way back to the ancient Greeks who discovered magnetite, which had magnetic properties.


== See also ==
== See also ==
[http://www.physicsbook.gatech.edu/Magnetic_Dipole Magnetic Dipole]
[http://www.physicsbook.gatech.edu/Magnetic_Dipole Magnetic Dipole]


===External links===
===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_charge Electric Charge]
* [https://en.wikipedia.org/wiki/Electric_dipole_moment Electric Dipole Moment]
 
* [https://en.wikipedia.org/wiki/Dipole Dipole]
[https://en.wikibooks.org/wiki/Physics_Exercises/Electrostatics Additional Dipole Derivations]
* [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/diptor.html Electric Dipole Torque]
 
[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]
[http://education.jlab.org/qa/historymag_01.html Magnet History]
* [https://en.wikipedia.org/wiki/Bohr_model Bohr Model]
 
* [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/diph2o.html Electric Dipole]
[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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