Divergence of an Electric Field due to an ideal dipole

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Homework Statement:

Am I tackling this problem in a correct way and if, is there any other way?

Relevant Equations:

##\nabla \times \vec E = \nabla \times (-\nabla \phi) = 0##
Given $$\vec E = -\nabla \phi$$ there $$\vec d \rightarrow 0, \phi(\vec r) = \frac {\vec p \cdot \vec r} {r^3}$$ and ##\vec p## is the dipole moment defined as $$\vec p = q\vec d$$

It's quite trivial to show that ##\nabla \times \vec E = \nabla \times (-\nabla \phi) = 0##. However, I want to show that ##\nabla \cdot \vec E = - \nabla^2 \phi = 0##.
My assumption of this equal ##0##, is from Gauss Law $$ \iint_{\partial V} \vec E \cdot d\vec S = \iiint_V \nabla \cdot \vec E \, dV = \frac {Q_{encl}}
{\epsilon_{0}}$$

However, when ##\vec d \rightarrow 0## we can't enclose one of the charges without enclosing the other one with any surface, hence ##\Sigma Q = 0##,
thus ##Q_{encl} = 0 \leftrightarrow \nabla \cdot \vec E = 0##

My questions: Am I reasoning correct, and does it exist any other way to show it? I'd rather not do a bruteforce calculation of ##- \nabla^2 \phi ##. =D
 

Answers and Replies

  • #2
Orodruin
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Clearly, as you have concluded, the divergence must be zero everywhere due to Gauss' law (at least away from the dipole, where things get more complicated than that - but that complication is most likely beyond the scope of your class and the problem formulation). The question becomes what you are allowed to assume and how strict you need to make your argument. For example, if you can start from the potential of each individual charge and the corresponding point-charge potential and use that to argue for the potential of the dipole, then that should be sufficient.

There is also nothing wrong with the brute force calculation, which is rather straightforward in this case given some basic identities regarding the derivatives of ##\vec r## and ##r##.
 
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There is also nothing wrong with the brute force calculation, which is rather straightforward in this case given some basic identities regarding the derivatives of ##\vec r## and ##r##.
Through calculations, given ##\vec P## is constant, can you assign ##\vec P## the values ##\vec P = (P_{1}, P_{2}, P_{3})## and regard the components as constants?
 
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Through calculations, given ##\vec P## is constant, can you assign ##\vec P## the values ##\vec P = (P_{1}, P_{2}, P_{3})## and regard the components as constants?
Assuming you are writing the components in a Cartesian basis, fine. However, it should be sufficient to know that ##\vec p## is a constant vector, you do not really need to assign coordinates unless you need to derive the derivatives of the position vector ##\vec r## and the distance ##r##. Of course, you can assign a Cartesian basis if that makes you happier. The final result will not depend on the choice of basis.
 
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Assuming you are writing the components in a Cartesian basis, fine. However, it should be sufficient to know that ##\vec p## is a constant vector, you do not really need to assign coordinates unless you need to derive the derivatives of the position vector ##\vec r## and the distance ##r##. Of course, you can assign a Cartesian basis if that makes you happier. The final result will not depend on the choice of basis.
Im quite confused how to derive the dot product. Through the quotient rule I obtain $$\nabla \phi_{x} = \frac {(\partial_{x} (\vec P \cdot \vec r))r^3) - (\vec P \cdot \vec r)(\partial_{x}r^3)} {r^6} $$

Which, makes it complicated to execute the Lapace operator $$-\nabla^2 \phi = -\partial_{i} \partial_{i} \phi$$ for the three components. Am I missing something?
 
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Perhaps it is not the best idea to look at particular components. Instead, try to focus on how the gradient and divergence act on ##\phi##. Also, Cartesian coordinates may or may not be the best choice for computing the Laplace operator acting on the dipole field (and that is about as much as I am comfortable saying here without giving too much away).
 
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  • #7
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I think I got it, defining ##0## can sometimes be tricky. Thanks for great answers, MB!
 

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