Find the Electric field due to an electric dipole at the origin

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SUMMARY

The discussion focuses on calculating the electric field due to an electric dipole at the origin using vector calculus. The key operation involves expanding the dot product of the dipole moment vector \(\vec{p}\) with the gradient operator \(\vec{\nabla}\) applied to the position vector \(\vec{r}\). The final expression for the electric field \(\vec{E}\) is derived as \(\vec{E} = -\vec{\nabla}\psi\), where \(\psi\) is the potential function related to the dipole. The participants confirm the calculations leading to the correct formulation of the electric field.

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  • Understanding of electric dipoles and their properties
  • Familiarity with vector calculus, particularly gradient operations
  • Knowledge of potential functions in electrostatics
  • Proficiency in using mathematical notation for derivatives and dot products
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MatinSAR
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Homework Statement
Find electric field due to dipole at any point using ##\vec E=-\nabla \phi##.
Relevant Equations
##\vec E=-\nabla \phi##
Question :

1699296919913.png


I have tried to solve but I struggle with this part:
1699297291071.png

Any help would be appreciated.
 

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Just expand the dot product and ##\vec r## then take derivatives.$$(\vec p\cdot \vec{\nabla})\vec r = \left(p_x\frac{\partial}{\partial x}+p_y\frac{\partial}{\partial y}+p_z\frac{\partial}{\partial z}\right)(x~\hat x+y~\hat y+z~\hat z)$$
 
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kuruman said:
Just expand the dot product and ##\vec r## then take derivatives.$$(\vec p\cdot \vec{\nabla})\vec r = \left(p_x\frac{\partial}{\partial x}+p_y\frac{\partial}{\partial y}+p_z\frac{\partial}{\partial z}\right)(x~\hat x+y~\hat y+z~\hat z)$$
Then it is equal to ##Pr^{-3}##, Am I right?!
 
Show me the math.
 
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kuruman said:
Show me the math.
1699298882251.png

Sorry that ##r^{-3}## was related to another part.
 
I have used:
##p_x\frac{\partial}{\partial x}y=0##
##p_x\frac{\partial}{\partial x}z=0##
##p_y\frac{\partial}{\partial y}x=0##
##p_y\frac{\partial}{\partial y}z=0##
##p_z\frac{\partial}{\partial z}x=0##
##p_z\frac{\partial}{\partial z}y=0##

##p_x\frac{\partial}{\partial x}x=p_x##
##p_y\frac{\partial}{\partial y}y=p_y##
##p_z\frac{\partial}{\partial z}z=p_z##
 
MatinSAR said:
I have used:
##p_x\frac{\partial}{\partial x}y=0##
##p_x\frac{\partial}{\partial x}z=0##
##p_y\frac{\partial}{\partial y}x=0##
##p_y\frac{\partial}{\partial y}z=0##
##p_z\frac{\partial}{\partial z}x=0##
##p_z\frac{\partial}{\partial z}y=0##

##p_x\frac{\partial}{\partial x}x=p_x##
##p_y\frac{\partial}{\partial y}y=p_y##
##p_z\frac{\partial}{\partial z}z=p_z##
Yes. What is your final answer when you put it all together?
 
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kuruman said:
Yes. What is your final answer when you put it all together?
##\vec P## , I guess.
 
Sorry, not that. I meant putting together the final expression ##\vec E=-\vec{\nabla}\psi=?##
 
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  • #10
kuruman said:
Sorry, not that. I meant putting together the final expression ##\vec E=-\vec{\nabla}\psi=?##
I am trying to solve ... I will send the work.

Thanks again for your help Prof.Kuruman🙏.
 
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  • #11
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  • #12
That's it. Good job!
 
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  • #13
kuruman said:
That's it. Good job!
Thanks a lot! Have a good day.
 
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