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

Click For Summary

Homework Help Overview

The discussion revolves around finding the electric field due to an electric dipole at the origin. Participants are exploring the mathematical expressions and derivatives involved in this context.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss expanding the dot product and taking derivatives, with some questioning the correctness of their expressions. There are attempts to clarify the steps involved in deriving the electric field.

Discussion Status

Several participants are engaged in exploring the mathematical details, with some providing partial expressions and seeking confirmation on their reasoning. There is an ongoing exchange of ideas without a clear consensus on the final expression.

Contextual Notes

Some participants express uncertainty regarding the final answer and the relationship between the derived expressions and the electric field. There are indications of confusion about specific terms and their implications in the context of the problem.

MatinSAR
Messages
673
Reaction score
204
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.
 

Attachments

  • 1699296851085.png
    1699296851085.png
    9.5 KB · Views: 108
Physics news on Phys.org
Just expand the dot product and ##\vec r## then take derivatives.$$(\vec p\cdot \vec{\nabla})\vec r = \left(p_x\frac{\partial}{\partial x}+p_y\frac{\partial}{\partial y}+p_z\frac{\partial}{\partial z}\right)(x~\hat x+y~\hat y+z~\hat z)$$
 
  • Like
Likes   Reactions: MatinSAR
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.
 
  • Like
Likes   Reactions: MatinSAR
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?
 
  • Like
Likes   Reactions: MatinSAR
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=?##
 
  • Like
Likes   Reactions: MatinSAR
  • #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🙏.
 
  • Like
Likes   Reactions: kuruman
  • #11
  • Like
Likes   Reactions: BvU
  • #12
That's it. Good job!
 
  • Love
Likes   Reactions: MatinSAR
  • #13
kuruman said:
That's it. Good job!
Thanks a lot! Have a good day.
 
  • Like
Likes   Reactions: kuruman

Similar threads

  • · Replies 10 ·
Replies
10
Views
1K
Replies
14
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
2
Views
3K
Replies
19
Views
4K
Replies
4
Views
3K
Replies
7
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K