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Homework Help: Electric Dipoles

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data
    An point electric dipole is located a vertical distance h above an infinite plane. What is the horizontal distance d from the dipole such that the electric field is zero at the horizontal distance d from the point dipole?

    2. Relevant equations

    [tex] V_{dip}=\frac{2p\cos{\theta}}{4\pi\epsilon_0 r^3}\hat{r}+\frac{p\sin{\theta}}{4\pi\epsilon_0 r^3}\hat{\theta} [/tex]

    3. The attempt at a solution

    There is an image charge a distance h below the plane, with same magnitude but opposite direction as that of the original dipole. Therefore, I got:

    [tex] E_x=\frac{-2p\cos{\theta}\sin{\theta}}{4\pi\epsilon_0 r^3}-\frac{-2p\cos{\theta}\sin{\theta}}{4\pi\epsilon_0 r^3}+\frac{-p\sin{\theta}\cos{\theta}}{4\pi\epsilon_0 r^3}+\frac{p\sin{\theta}\cos{\theta}}{4\pi\epsilon_0 r^3} [/tex]

    But it looks as if the E_x component is always finite. I know the final answer is [tex] d=\sqrt{2}h [/tex] but I can't seem to figure out how to get the answer. The diagram that I have constructed for the question is attached.
     

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  3. Feb 20, 2010 #2

    gabbagabbahey

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    Surely, this is supposed to be

    [tex]\textbf{E}_{dip}=\frac{2p\cos{\theta}}{4\pi\epsilon_0 r^3}\hat{r}+\frac{p\sin{\theta}}{4\pi\epsilon_0 r^3}\hat{\theta} [/tex]

    Right?

    In any case, this gives the electric field due to a dipole located at the origin, oriented at an angle [itex]\theta[/itex] to the positive z-axis.....Your dipole (and its image) is not at the origin here.


    Why are you only looking at the x-component? If the electric field is zero at a given point, all of its components are zero there.

    With the way your problem statement is written, I would interpret it as meaning the point at which [itex]\textbf{E}=0[/itex] is at [itex]x=d[/itex] and [itex]z=h[/itex], not [itex]z=0[/itex] (Assuming you are using the z-axis as the vertical and the x-axis as the horizontal)

    In any case, it might be easier for you to find the correct expression for the electric field if you start by fining the potential.
     
  4. Feb 20, 2010 #3
    Yes it was a mistake, it was supposed to be [tex] E_{dip} [/tex]

    I was trying to determine if the x-component cancels before moving on to the y. But I was stuck.

    I don't get you about the origin. It seems as though I have considered the origin to be at the point of the dipoles when computing their contribution to the electric field. The [tex] r [/tex] in all the terms can be expressed as [tex] r=\sqrt{h^2+d^2} [/tex] and so the last 2 terms cancel and I'm left with a finite x-component. Btw, I think I made a mistake in the expression, it should be [tex]
    E_x=\frac{-2p\cos{\theta}\sin{\theta}}{4\pi\epsilon_0 r^3}+\frac{-2p\cos{\theta}\sin{\theta}}{4\pi\epsilon_0 r^3}+\frac{-p\sin{\theta}\cos{\theta}}{4\pi\epsilon_0 r^3}+\frac{p\sin{\theta}\cos{\theta}}{4\pi\epsilon _0 r^3}
    [/tex]

    so this was the expression I was stuck at.
    Thanks for your time
     
    Last edited: Feb 20, 2010
  5. Feb 21, 2010 #4

    gabbagabbahey

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    Sure, [itex]r[/itex] is the same for both the dipole and its image (going off of your diagram), but what about [itex]\theta[/itex], [itex] \mathbf{\hat{r}}[/tex] and [tex]\mathbf{\hat\theta}}[/itex]?
     
  6. Feb 21, 2010 #5
    well, [tex] \theta [/tex] is the same since both triangles are similar, and the directions of [tex] \hat{r} [/tex] and [tex] \hat{\theta} [/tex] are taken into account in expressing [tex] E_x [/tex]. Is this right?
     
  7. Feb 21, 2010 #6

    gabbagabbahey

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    No, the vector from the first dipole to the field point makes an angle greater than 90 degrees with the positive z-axis, while the vector from the image dipole to the field point makes an angle smaller than 90 degrees with the positive z-axis. (You've incorrectly labeled [itex]\theta[/itex] for the original dipole).

    The expressions you seem to be using for [itex]\hat{\mathbf{r}}[/tex] and [tex]\hat{\mathbf{\theta}}[/tex] don't take into account that [itex]\textbf{r}[/itex] isn't measured from the origin, but rather from the locations of the dipoles (there are two different [itex]\textbf{r}[/itex]'s)

    Also, why are you assuming the the (real) dipole is oriented perpendicular to the conducting plane? Was this part of your original problem statement?

    Again, you will probably have a much easier time finding the correct field by first finding the potential...place a dipole (pointing in the positive z-direction) at [itex](0,0,h)[/itex] and another oppositely oriented image dipole at [itex](0,0,-h)[/itex]...what is the potential due to this arrangement at an arbitrary point in space [itex](x,y,z)[/itex]?
     
  8. Feb 21, 2010 #7
    oh yes the dipole moment points upwards, sorry for leaving that out.

    I skipped a few steps, the first dipole makes an angle larger than 90 degrees. So the cosine of the angle gives the negative times the cosine of theta in the diagram. If the first dipole points up, the image dipole should point downwards and so i got the negative expression again for the image dipole.

    Hmm, I've placed my origin at the first dipole, calculated the electric field from that, then moved my origin to the second dipole and calculated the electric field from that one.

    If I calculated the potential, how would I know at which point of the plane would the electric field be zero?
     
  9. Feb 21, 2010 #8

    gabbagabbahey

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    Okay, I see what you are doing now. You still have a slight error, using this method, you should end up with

    [tex]E_x=\frac{-3p\sin\theta\cos\theta}{2\pi\epsilon_0 r^3}[/tex]

    (Your last term has a sign error)

    Now, given that [itex]\sin\theta=\frac{d}{r}[/itex] and [itex]\cos\theta=\frac{h}{r}[/itex], you can see that this expression only vanishes for [itex]d=0[/itex]....So, what went wrong?....Does your image dipole really point opposite to the real dipole (pretend it's made of two point charges)?

    [itex]\textbf{E}=-\mathbf{\nabla}V[/itex], so the gradient of the potential must be zero at points where the electric field vanishes.

    Calculating the potential first has two advantages:

    (1) It allows you to make sure you've correctly set-up your image problem by substituting [itex]z=0[/itex] into your expression and checking that you do indeed get zero.

    (2) It allows you to determine all 3 components of the electric field at once, just by taking the negative gradient of a simple scalar function.
     
    Last edited: Feb 21, 2010
  10. Feb 21, 2010 #9
    hmm, the question did ask specifically for a point on the surface. But they said the surface was conducting. Is the placing of the image I have done only for grounded surfaces?
     
  11. Feb 21, 2010 #10

    gabbagabbahey

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    See my edited post. It seems they want you to consider the point to be infinitesimally close to the surface, so there's nothing wrong with assuming z=0. (Technically, you should take a general expression for the field in the region [itex]z>0[/itex] and take the limit as [itex]z[/itex] approaches zero from the positive direction. But it seems that your instructor simply intends you to simply pretend that the electric field you calculate via MOI is valid at [itex]z=0[/itex], despite there being a conductor there)
     
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