Components of an Electric field due to a dipole

In summary: Well, at least you are on the right track now. In summary, the problem is that you need to find the electric fields at distant points, but you are struggling with simplifying the numerator and denominator.
  • #1
forensics409
6
0

Homework Statement


The problem is: Show that the components of [tex]\vec{E}[/tex] due to a dipole are given at distant points, by Ex=[tex]\frac{1}{4\pi\epsilon{o}}[/tex] [tex]\frac{3pxz}{(x^2+z^2)^{5/2}}[/tex] and Ez=[tex]\frac{1}{4\pi\epsilon{o}}[/tex] [tex]\frac{p(2z^2-x^2)}{(x^2+z^2)^(\frac{5}{2})}}[/tex]


http://physweb.bgu.ac.il/COURSES/PHYSICS2_B/2009A/homework/Homework-2_files/image006.jpg

Homework Equations



E=[tex]\frac{1}{4\pi\epsilon{o}}[/tex] [tex]\frac{Q}{r^2}[/tex]
p=qd

The Attempt at a Solution



I have tried to break the fields of each one into vector components and add the components, however, it got really messy really quickly and after simplifying it a bit i got a ridiculous equation for just the x component, i had no clue where to go and gave up on even try to get the z component.

Ex=[tex]\frac{q}{4\pi\epsilon{o}}[/tex][tex]\frac{(x^2+(z+\frac{d}{2})^{2})^{\frac{3}{2}}-(x^2+(z-\frac{d}{2})^{2})^{\frac{3}{2}}}{((x^{2}+z^{2})^{2} + (\frac{d^{2}x^{2}}{2}-\frac{d^{2}z^{2}}{2}+\frac{d^4}{16}))^{\frac{3}{2}}}[/tex]
 
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  • #2
I haven't checked your expression, but the key thing to notice here is that you are asked to find [itex]E_x[/itex] and [itex]E_z[/itex] at distant points. To me, that means you should assume that the distance to the field point from the center of the dipole is much greater than the length of the dipole, or

[tex]\frac{d}{\sqrt{x^2+z^2}}\ll 1[/tex]
 
  • #3
But how does that help?
 
  • #4
forensics409 said:
But how does that help?

Well, looking at the denominator of your expression; if you factor out an [tex](x^2+z^2)^3[/itex], you're left with [tex](1+\text{some stuff})^{3/2}[/itex]. It shouldn't take you much effort to show that the "some stuff" is much smaller than one at distant points, and so your denominator is approximately just [tex](x^2+z^2)^3[/itex].
 
  • #5
Ok. but my biggest problem is then simplifying the numerator, and I tried using binomial expansion for that and you get (x[tex]^{2}[/tex]+z[tex]^{2}[/tex]+dz)[tex]^{3/2}[/tex] - (x[tex]^{2}[/tex]+z[tex]^{2}[/tex]-dz)[tex]^{3/2}[/tex] and a denominator of (x[tex]^{2}[/tex] + z[tex]^{2}[/tex])[tex]^{3}[/tex]

sorry to ask but what do you do from there?
 
  • #6
To be honest, your expression just doesn't look right to me. And I think you are making the calculations more difficult than they need to be by working in Cartesian coordinates.

If I were you, I'd redo your calculation using polar coordinates, [itex]x=r\cos\theta[/itex] and [itex]z=r\sin\theta[/itex]. The advantage of this is that your "at distant points" approximation simply means that [itex]\frac{d}{r}\ll 1[/itex], and so writing your expression in terms of the dimensionless variable [itex]\eta\equiv\frac{d}{r}[/itex], you'll only need to Taylor expand it around the point [itex]\eta=0[/itex] and keep terms up to first order.
 
  • #7
actually, doing the whole thing from the electric field is ridiculously stupid. I realized I should start it in Volts so it looks like this

V = [tex]\frac{q}{4\pi\epsilon}[/tex] ([tex]\frac{1}{r-\frac{l}{2}cos\vartheta}[/tex]-[tex]\frac{1}{r+\frac{l}{2}cos\vartheta}[/tex]) which then simplifies to V=[tex]\frac{q}{4\pi\epsilon}[/tex][tex]\frac{lcos\vartheta}{r^2}[/tex]

V=[tex]\frac{pz}{(r^3)4\pi\epsilon}[/tex]

then -[tex]\nabla[/tex]V=E so then you just take the partial derivatives of it in terms of x and z.

I feel stupid for not thinking of it sooner
 

FAQ: Components of an Electric field due to a dipole

What is an electric dipole?

An electric dipole is a pair of equal and opposite charges separated by a small distance. This creates a dipole moment, which is a measure of the strength and direction of the dipole.

What are the components of an electric field due to a dipole?

The components of an electric field due to a dipole are the parallel component and the perpendicular component. The parallel component is directed along the axis of the dipole, while the perpendicular component is directed perpendicular to the axis.

How do you calculate the electric field due to a dipole?

The electric field due to a dipole can be calculated using the formula E = (1/4πε0) * (2p/r3), where p is the dipole moment, r is the distance from the dipole, and ε0 is the permittivity of free space.

What is the direction of the electric field due to a dipole?

The direction of the electric field due to a dipole is always from the positive charge to the negative charge. This means that the electric field lines will point away from the positive charge and towards the negative charge.

How does the strength of the electric field due to a dipole change with distance?

The strength of the electric field due to a dipole decreases with distance from the dipole. This is because the electric field follows an inverse square law, meaning that it decreases with the square of the distance from the source.

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