- #1
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!