Electric Dipole

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

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.

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

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

We now have values for [math]\displaystyle{ d, q, \theta_+, \theta_-, \vec r_+, \vec r_- }[/math]. This is enough to calculate [math]\displaystyle{ E_{net} }[/math] in both directions. The general formula for electric field strength from a Point Charge is [math]\displaystyle{ E = \frac{1}{4\pi\epsilon_0} \frac{q}{|\vec r|^2} \hat r }[/math]. Then [math]\displaystyle{ |E_+| = \frac{1}{4\pi\epsilon_0} \frac{q_+}{|\vec r_+|^2} }[/math] and [math]\displaystyle{ |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]\displaystyle{ r_+, r_- }[/math] to the form in the denominator of the sine and cosine. We will use this later.

First we calculate [math]\displaystyle{ E_{net_y} }[/math]. [math]\displaystyle{ 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]\displaystyle{ q_+ = -q_- }[/math]. [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]

Now it gets ugly, we expand our radii and sines. To recap, [math]\displaystyle{ sin(\theta_+) = \frac{p_y}{\sqrt{(p_x - \frac{d}{2})^2+p_y^2}} }[/math], [math]\displaystyle{ sin(\theta_-) = \frac{p_y}{\sqrt{(p_x + \frac{d}{2})^2+p_y^2}} }[/math], [math]\displaystyle{ |r_+| = \sqrt{(p_x - \frac{d}{2})^2 +p_y^2} }[/math] and [math]\displaystyle{ |r_-| = \sqrt{(p_x + \frac{d}{2})^2 +p_y^2} }[/math], giving us

[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]

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]\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] 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]\displaystyle{ sin(\theta_+) = \frac{p_y}{\sqrt{(p_x - \frac{d}{2})^2+p_y^2}} }[/math], [math]\displaystyle{ 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]\displaystyle{ \theta_- }[/math], the result is that

[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 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]\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]. Since we are on the parallel axis, we know that [math]\displaystyle{ E_{net_y} = 0 }[/math], and [math]\displaystyle{ p_y = 0 }[/math].

Simplifies to

[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 \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]\displaystyle{ 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]\displaystyle{ \frac{d}{2} }[/math] with [math]\displaystyle{ a }[/math].

[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) = \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]\displaystyle{ p_x \gt \gt a }[/math], we can assume that [math]\displaystyle{ p_x^2 + a^2 }[/math] is very close to [math]\displaystyle{ p_x^2 }[/math]. Then

[math]\displaystyle{ 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]\displaystyle{ E_{net_y} = 0 }[/math] because the vertical forces from both point charges cancel, leaving only horizontal forces.

[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]

In this case though, [math]\displaystyle{ p_x = 0 }[/math]

[math]\displaystyle{ 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]\displaystyle{ a = \frac{d}{2} }[/math].

[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{-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]\displaystyle{ d }[/math] is much smaller than [math]\displaystyle{ p_y }[/math], [math]\displaystyle{ a }[/math] is also small, so we can assume that the denominator is just [math]\displaystyle{ p_y }[/math]. This allows us to simplify the resulting equation to

[math]\displaystyle{ E_{net_x} \approx \frac{q_+}{4\pi\epsilon_0} \frac{-2a}{p_y^3} }[/math]

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Additional Dipole Derivations

Electric Dipole Moment

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