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MatinSAR
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Then it is equal to ##Pr^{-3}##, Am I right?!kuruman said:Just expand the dot product and ##\vec r## then take derivatives.$$(\vec p\cdot \vec{\nabla})\vec r = \left(p_x\frac{\partial}{\partial x}+p_y\frac{\partial}{\partial y}+p_z\frac{\partial}{\partial z}\right)(x~\hat x+y~\hat y+z~\hat z)$$
kuruman said:Show me the math.
Yes. What is your final answer when you put it all together?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##
##\vec P## , I guess.kuruman said:Yes. What is your final answer when you put it all together?
I am trying to solve ... I will send the work.kuruman said:Sorry, not that. I meant putting together the final expression ##\vec E=-\vec{\nabla}\psi=?##
Thanks a lot! Have a good day.kuruman said:That's it. Good job!
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.
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.
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.
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.
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.