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

  • #1
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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  • #2
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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  • #3
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?!
 
  • #4
Show me the math.
 
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  • #5
kuruman said:
Show me the math.
1699298882251.png

Sorry that ##r^{-3}## was related to another part.
 
  • #6
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##
 
  • #7
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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  • #8
kuruman said:
Yes. What is your final answer when you put it all together?
##\vec P## , I guess.
 
  • #9
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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1. What is an electric dipole?

An electric dipole is a pair of equal and opposite charges separated by a small distance. This creates a dipole moment, which is a measure of the strength of the dipole.

2. How is the electric field due to a dipole calculated?

The electric field due to a dipole at a point is calculated by taking the difference of the electric field vectors at that point due to each individual charge in the dipole.

3. What is the direction of the electric field due to a dipole?

The direction of the electric field due to a dipole is along the line connecting the two charges and away from the positive charge towards the negative charge.

4. How does the distance from the dipole affect the magnitude of the electric field?

The magnitude of the electric field due to a dipole decreases with distance from the dipole. It follows an inverse square law, meaning that the field strength decreases as the distance squared.

5. Can the electric field due to a dipole be zero?

Yes, the electric field due to a dipole can be zero at certain points along the line connecting the two charges. These points are known as the equatorial points and are located at a distance of half the dipole separation from the center of the dipole.

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