Electric field of a dipole

  • #1
912
19
Hi

I have a question about the electric field of a dipole located at the origin. We know that the
coordinate free form for the electric field of a dipole is

[tex]\vec{\mathbf{E}}=\frac{1}{4\pi \epsilon_o}\frac{1}{r^3}\left[\frac{3(\vec{p}\cdot \vec{r})\vec{r}}{r^2}-\vec{p}\right] [/tex]

Now let

[tex]\vec{r}= x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}} [/tex]

be the vector of some point in space. Assume that [itex] \vec{p}=p(\hat{\mathbf{z}})[/itex].

So

[tex] \vec{p}\cdot \vec{r} = zp [/tex]

so plugging everything we get for the electric field,

[tex]\vec{\mathbf{E}}=\frac{p}{4\pi \epsilon_o}\left[\frac{3z(x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}})-(x^2+y^2+z^2)\hat{\mathbf{z}}}{(x^2+y^2+z^2)^{5/2}}\right] [/tex]

Now, if I want to get the electric field on the negative z axis, then I put x=y=0 and
z= -r , so I get

[tex]\vec{\mathbf{E}}=\frac{2p(-r)^2 \hat{\mathbf{z}}}{4\pi \epsilon_o(-r)^5} [/tex]

which is after evaluating the brackets,

[tex]\vec{\mathbf{E}}=-\frac{2p \hat{\mathbf{z}}}{4\pi \epsilon_o r^3} [/tex]

which points in the negative z direction, but if we look at the original formula for the
coordinate form of the electric field and then evaluate the electric field on the negative
z axis, it points toward the positive z direction. So whats happening here ?

thanks
 

Answers and Replies

  • #2
311
1
Your fallacy lies in evaluating the denominator. Note the following

[tex]\left(x^2\right)^{1/2} = |x|[/tex]

and NOT x. That fixes your problem.
 
  • #3
912
19
thanks prahar

how could i miss that ? :tongue:
 

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