Physics Book - User contributions [en]

Electric Dipole

2015-12-05T08:40:22Z

Jmorton32: /* Examples */

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 [[User:Jmorton32|Jmorton32]] ([[User talk:Jmorton32|talk]]) 02:52, 19 October 2015 (EDT)

==Mathematical Models==

===An Exact Model===
[[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>.

<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>.

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.

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>.

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>

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

<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>

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.

<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> 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

<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 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==
We can simplify the solution for many cases

===On the Parallel Axis===
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

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

=

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

<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===
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>
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>.

<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{-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==

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

===Simple===

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.

<div class="toccolours mw-collapsible mw-collapsed">
===Click for Solution===
<div class="mw-collapsible-content">
The center of the dipole is at the origin and 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.

<math>1.6\times 10^{-19} \times 9 \times 10^9
\frac{-9 \times 10^{-10}}
{3 \times 10^{-8^3}} = -48000 \frac{N}{C}</math> on the y axis, as a vector: <math><0, -48000, 0></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>.
</div>
</div>

===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) calculate the charge on the ping-pongball to levitate the dipole

b) the dipole is turned 90 degrees clockwise, without changing its position relative to the ball, what effect does this have on the ball?

<div class="toccolours mw-collapsible mw-collapsed">

===Click for Solutions===
<div class="mw-collapsible-content">
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>

Solving for <math>|Q|</math>: <math>|Q| = \Bigg(\frac{1}{4\pi\epsilon_0}\Bigg)^{-1} \frac{mgL^3}{2p}</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>

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.

</div>
</div>

==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.

==History==

Electric dipoles have been understood since the mid to late 1800s.

== See also ==

===External links===

[https://en.wikibooks.org/wiki/Physics_Exercises/Electrostatics Additional Dipole Derivations]

[https://en.wikipedia.org/wiki/Electric_dipole_moment Electric Dipole Moment]

==References==

[[Category:Fields]]

Electric Dipole

2015-12-05T08:19:30Z

Jmorton32: /* Middling */

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 [[User:Jmorton32|Jmorton32]] ([[User talk:Jmorton32|talk]]) 02:52, 19 October 2015 (EDT)

==Mathematical Models==

===An Exact Model===
[[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>.

<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>.

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.

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>.

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>

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

<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>

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.

<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> 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

<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 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==
We can simplify the solution for many cases

===On the Parallel Axis===
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

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

=

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

<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===
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>
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>.

<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{-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==

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

===Simple===

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.

<div class="toccolours mw-collapsible mw-collapsed">
===Click for Solution===
<div class="mw-collapsible-content">
The center of the dipole is at the origin and 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.

<math>1.6\times 10^{-19} \times 9 \times 10^9
\frac{-9 \times 10^{-10}}
{3 \times 10^{-8^3}} = -48000 \frac{N}{C}</math> on the y axis, as a vector: <math><0, -48000, 0></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>.
</div>
</div>

===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) calculate the charge on the ping-pongball to levitate the dipole

b) the dipole is turned 90 degrees clockwise, without changing its position relative to the ball, what effect does this have on the ball?

<div class="toccolours mw-collapsible mw-collapsed">

===Click for Solutions===
<div class="mw-collapsible-content">
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>

Solving for <math>|Q|</math>: <math>|Q| = \Bigg(\frac{1}{4\pi\epsilon_0}\Bigg)^{-1} \frac{mgL^3}{2p}</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>

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.

</div>
</div>

===Difficult===

==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.

==History==

Electric dipoles have been understood since the mid to late 1800s.

== See also ==

===External links===

[https://en.wikibooks.org/wiki/Physics_Exercises/Electrostatics Additional Dipole Derivations]

[https://en.wikipedia.org/wiki/Electric_dipole_moment Electric Dipole Moment]

==References==

[[Category:Fields]]

Electric Dipole

2015-12-05T08:19:18Z

Jmorton32: /* Middling */

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 [[User:Jmorton32|Jmorton32]] ([[User talk:Jmorton32|talk]]) 02:52, 19 October 2015 (EDT)

==Mathematical Models==

===An Exact Model===
[[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>.

<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>.

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.

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>.

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>

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

<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>

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.

<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> 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

<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 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==
We can simplify the solution for many cases

===On the Parallel Axis===
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

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

=

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

<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===
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>
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>.

<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{-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==

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

===Simple===

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.

<div class="toccolours mw-collapsible mw-collapsed">
===Click for Solution===
<div class="mw-collapsible-content">
The center of the dipole is at the origin and 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.

<math>1.6\times 10^{-19} \times 9 \times 10^9
\frac{-9 \times 10^{-10}}
{3 \times 10^{-8^3}} = -48000 \frac{N}{C}</math> on the y axis, as a vector: <math><0, -48000, 0></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>.
</div>
</div>

===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) calculate the charge on the ping-pongball to levitate the dipole
b) the dipole is turned 90 degrees clockwise, without changing its position relative to the ball, what effect does this have on the ball?

<div class="toccolours mw-collapsible mw-collapsed">
===Click for Solutions===
<div class="mw-collapsible-content">
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>

Solving for <math>|Q|</math>: <math>|Q| = \Bigg(\frac{1}{4\pi\epsilon_0}\Bigg)^{-1} \frac{mgL^3}{2p}</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>

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.

</div>
</div>

===Difficult===

==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.

==History==

Electric dipoles have been understood since the mid to late 1800s.

== See also ==

===External links===

[https://en.wikibooks.org/wiki/Physics_Exercises/Electrostatics Additional Dipole Derivations]

[https://en.wikipedia.org/wiki/Electric_dipole_moment Electric Dipole Moment]

==References==

[[Category:Fields]]

Electric Dipole

2015-12-05T07:57:58Z

Jmorton32: /* Simple */

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 [[User:Jmorton32|Jmorton32]] ([[User talk:Jmorton32|talk]]) 02:52, 19 October 2015 (EDT)

==Mathematical Models==

===An Exact Model===
[[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>.

<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>.

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.

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>.

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>

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

<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>

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.

<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> 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

<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 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==
We can simplify the solution for many cases

===On the Parallel Axis===
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

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

=

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

<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===
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>
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>.

<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{-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==

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

===Simple===

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.

<div class="toccolours mw-collapsible mw-collapsed">
===Click for Solution===
<div class="mw-collapsible-content">
The center of the dipole is at the origin and 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.

<math>1.6\times 10^{-19} \times 9 \times 10^9
\frac{-9 \times 10^{-10}}
{3 \times 10^{-8^3}} = -48000 \frac{N}{C}</math> on the y axis, as a vector: <math><0, -48000, 0></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>.
</div>
</div>

===Middling===
===Difficult===

==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.

==History==

Electric dipoles have been understood since the mid to late 1800s.

== See also ==

===External links===

[https://en.wikibooks.org/wiki/Physics_Exercises/Electrostatics Additional Dipole Derivations]

[https://en.wikipedia.org/wiki/Electric_dipole_moment Electric Dipole Moment]

==References==

[[Category:Fields]]

Electric Dipole

2015-12-05T07:36:26Z

Jmorton32: /* History */

Electric Dipole

2015-12-05T01:36:35Z

Jmorton32: /* References */

Electric Dipole

2015-12-05T01:36:05Z

Jmorton32: /* Connectedness */

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 [[User:Jmorton32|Jmorton32]] ([[User talk:Jmorton32|talk]]) 02:52, 19 October 2015 (EDT)

==Mathematical Models==

===An Exact Model===
[[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>.

<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>.

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.

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>.

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>

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

<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>

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.

<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> 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

<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 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==
We can simplify the solution for many cases

===On the Parallel Axis===
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

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

=

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

<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===
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>
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>.

<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{-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==

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

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

==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.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

===External links===

[https://en.wikibooks.org/wiki/Physics_Exercises/Electrostatics Additional Dipole Derivations]

[https://en.wikipedia.org/wiki/Electric_dipole_moment Electric Dipole Moment]

==References==

This section contains the the references you used while writing this page

[[Category:Fields]]

Electric Dipole

2015-12-04T23:56:45Z

Jmorton32: /* On the Perpendicular Axis */

Electric Dipole

2015-12-04T23:56:19Z

Jmorton32: /* External links */

Electric Dipole

2015-12-04T23:56:10Z

Jmorton32: /* See also */

Electric Dipole

2015-12-04T20:22:48Z

Jmorton32: /* On the Perpendicular Axis */

Electric Dipole

2015-12-04T19:54:52Z

Jmorton32: /* On the Perpendicular Axis */

Electric Dipole

2015-12-04T19:17:59Z

Jmorton32: /* On the Parallel Axis */

Electric Dipole

2015-12-04T19:17:42Z

Jmorton32: /* Special Cases */

Electric Dipole

2015-12-04T18:15:35Z

Jmorton32: /* On the Parallel Axis */

Electric Dipole

2015-12-04T17:52:33Z

Jmorton32: /* Special Cases */

Electric Dipole

2015-12-04T17:36:02Z