Electric Dipole: Difference between revisions
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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>. | 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>. | ||
We know that the net electric field at <math>p</math> is equal to the sum of all electric fields in our system, in this case <math>E_{net} = E_{q_-} + E_{q_+}</math>. Then we divide <math>E_{net}</math> into its x and y components and calculate the electric field for each. Let's do y first. | |||
<math>E_{net_y} = E_{q_{-_y}} + E_{q_{+_y}}</math>. The equation for electric field is <math>\frac{1}{4\pi\epsilon_0} \frac{q}{|\vec r|^2} \hat r</math>. Where <math>\hat r</math> is the unit vector in the direction from the the charge to the point p. We want only the component of this in the y direction, which means | |||
==Examples== | ==Examples== | ||
Revision as of 14:36, 3 December 2015
An Electric Dipole is a pair of equal and opposite Point Charges separated by a small distance. Electric dipoles have a number of interesting properties.
claimed by Jmorton32 (talk) 02:52, 19 October 2015 (EDT)
Mathematical Models
An Exact Model
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].
We know that the net electric field at [math]\displaystyle{ p }[/math] is equal to the sum of all electric fields in our system, in this case [math]\displaystyle{ E_{net} = E_{q_-} + E_{q_+} }[/math]. Then we divide [math]\displaystyle{ E_{net} }[/math] into its x and y components and calculate the electric field for each. Let's do y first.
[math]\displaystyle{ E_{net_y} = E_{q_{-_y}} + E_{q_{+_y}} }[/math]. The equation for electric field is [math]\displaystyle{ \frac{1}{4\pi\epsilon_0} \frac{q}{|\vec r|^2} \hat r }[/math]. Where [math]\displaystyle{ \hat r }[/math] is the unit vector in the direction from the the charge to the point p. We want only the component of this in the y direction, which means
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