MatinSAR
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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.
PREREQUISITESStudents and professionals in physics, particularly those specializing in electromagnetism, as well as anyone interested in the mathematical foundations of electric fields and dipole interactions.
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!