Electric Dipole: Difference between revisions

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===An Exact Model===
===An Exact Model===
[[File:Dipole.png|300px|thumb|An Electric Dipole]]
[[File:Dipole.png|300px|thumb|An Electric Dipole]]
Since an electric dipole is made up of 2 electric point charges, the electric field of the dipole can be calculated by summing the electric fields contributed by each point charge.  In the example, the field at point P, <math>E_{P}</math> is equivalent to the sum of the electric field from the positive charge <math>q+</math> and the negative charge <math>q-</math>.
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.
In other words, <math>E_{P} = E_{P_{q+}} + E_{P_{q-}}</math>.


Substituting in the equation for the electric field from a point charge we get <math>\vec E_{P} = \frac{1}{4\pi\epsilon_{0}} \times \frac{q_{+}}{|r_{+}|^{2}} \times \hat r_{+} + \frac{1}{4\pi\epsilon_{0}} \times \frac{q_{-}}{|r_{-}|^{2}} \times \hat r_{-}</math>.
Then we begin by calculating <math>r_+</math> and <math>r_-</math> the radii from the positive and negative particles to the point <math>p</math>. In this example, we will assume that the positive particle is closer to <math>p</math>, but its simple to modify this derivation for the opposite case. First, we divide <math>r</math> into its x and y components, <math> r_x = r * cos(\theta) </math> and <math> r_y = r * sin(\theta) </math>.
 
A simple refactoring gives <math>\vec E_{P} = \frac{1}{4\pi\epsilon_{0}} \times (\frac{q_{+}}{|r_{+}|^{2}}\hat r_{+} + \frac{q_{-}}{|r_{-}|^{2}} \hat r_{-})</math>.
 
Since <math>q_+ = -1 * q_-</math>, this is equivalent to <math>\vec E_{P} = \frac{q_+}{4\pi\epsilon_{0}} \times (\frac{\hat r_{+}}{|r_{+}|^{2}} - \frac{\hat r_{-}}{|r_{-}|^{2}})</math>
 
<math>|r_+| \text{ and } |r_-|</math> can then be calculated as <math>|r_+| = \sqrt{(|r_x| + \frac{d}{2})^2 + |r_y|^2}</math>, by decomposing <math>\vec r</math> into its components and factoring.
 
By a similar method, <math>|r_-| = \sqrt{(|r_x| - \frac{d}{2})^2 + |r_y|^2}</math>.
 
By substituting <math>|\vec r| \cos(\theta) \text{ and } |\vec r| \sin(\theta)</math> for  <math>|r_x| \text{ and } |r_y|</math> respectively, we get
 
<math>|r_+| = \sqrt{(|\vec r| \cos(\theta) + \frac{d}{2})^2 + (|\vec r| \sin(\theta))^2}</math> and <math>|r_-| = \sqrt{(|\vec r| \cos(\theta) - \frac{d}{2})^2 + (|\vec r| \sin(\theta))^2}</math>
 
We next expand the squared terms inside the radical: <math>\sqrt{(|\vec r| \cos(\theta))^2 + \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4} +(|\vec r| \sin(\theta))^2}</math>
 
Rearranging, <math>\sqrt{|\vec r|^2 * (\cos(\theta)^2 + \sin(\theta)^2) + \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}} = \sqrt{|\vec r|^2 + \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}} = |r_+|</math>
 
By a similar method, <math>|r_-| = \sqrt{|\vec r|^2 - \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}}</math>
 
Plugging this into the earlier equation, we get the monstrosity <math>E_{P} = \frac{q_+}{4\pi\epsilon_{0}} \times (\frac{\hat r_{+}}{|\vec r|^2 + \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}} - \frac{\hat r_{-}}{|\vec r|^2 - \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}})</math>
 
To subtract the fractions, we first put them in like terms:
 
<math> \frac{\hat r_{+}}{|\vec r|^2 + \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}}
\times
\frac{|\vec r|^2 - \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}}{|\vec r|^2 - \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4}})
=
\frac{\hat r_{+}\times (|\vec r|^2 - \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4})}{|\vec r|^4 - \frac{|\vec r|^2 d |\vec r| \cos(\theta)}{2} + \frac{|\vec r|^2 d^2}{4} + \frac{|\vec r|^2 d |\vec r| \cos(\theta)}{2} - \frac{d^2 |\vec r|^2 \cos(\theta)^2}{4} + \frac{d^3|\vec r|cos(\theta)}{8} + \frac{|\vec r|^2 d^2}{4} - \frac{d^3|\vec r|cos(\theta)}{8} + \frac{d^4}{16}}</math>
 
Cancelling like terms simplifies this to
<math>\frac{\hat r_{+}\times (|\vec r|^2 - \frac{d |\vec r| \cos(\theta)}{2} + \frac{d^2}{4})}{|\vec r|^4 + \frac{|\vec r|^2 d^2}{2} -\frac{d^2 |\vec r|^2 \cos(\theta)^2}{4} + \frac{d^4}{16}}</math>


==Examples==
==Examples==

Revision as of 15:13, 3 December 2015

An Electric Dipole is a pair of equal and opposite Point Charges separated by a small distance.

claimed by Jmorton32 (talk) 02:52, 19 October 2015 (EDT)

Mathematical Models

An Exact Model

An Electric Dipole

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 are of equal and opposite charge. We then wish to know the electric field due to the dipole at some point [math]\displaystyle{ p }[/math] in the plane (see the figure). [math]\displaystyle{ p }[/math] can be considered either a distance [math]\displaystyle{ [x_0, y_0] }[/math] from the midpoint of the dipole, or a distance [math]\displaystyle{ r }[/math] and an angle [math]\displaystyle{ \theta }[/math] as in the diagram.

Then we begin by calculating [math]\displaystyle{ r_+ }[/math] and [math]\displaystyle{ r_- }[/math] the radii from the positive and negative particles to the point [math]\displaystyle{ p }[/math]. In this example, we will assume that the positive particle is closer to [math]\displaystyle{ p }[/math], but its simple to modify this derivation for the opposite case. First, we divide [math]\displaystyle{ r }[/math] into its x and y components, [math]\displaystyle{ r_x = r * cos(\theta) }[/math] and [math]\displaystyle{ r_y = r * sin(\theta) }[/math].

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