Divergence of an Electric Field due to an ideal dipole

AI Thread Summary
The discussion focuses on deriving the divergence of the electric field due to an ideal dipole, starting from the potential function and applying Gauss's Law. It is established that the divergence of the electric field is zero because the enclosed charge is zero when the dipole moment approaches zero. Participants suggest that while a brute force calculation of the Laplacian is straightforward, understanding the potential from individual charges can simplify the argument. The conversation also touches on the appropriateness of using Cartesian coordinates for calculations and emphasizes that the final results are independent of the chosen basis. Overall, the reasoning about the divergence being zero is confirmed, with suggestions for alternative approaches to the derivation.
rakso
Messages
18
Reaction score
0
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
 
Physics news on Phys.org
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##.
 
Orodruin said:
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?
 
rakso said:
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.
 
  • Like
Likes rakso
Orodruin said:
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?
 
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).
 
  • Like
Likes rakso
I think I got it, defining ##0## can sometimes be tricky. Thanks for great answers, MB!
 
Back
Top