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Find the Electric field due to an electric dipole at the origin
- Thread starter MatinSAR
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In summary, to find the electric field due to an electric dipole at the origin, one must consider the dipole moment, defined as the product of charge and the separation distance between the charges. The electric field at a point in space can be derived using the formula for the electric field due to a dipole, which is dependent on the distance from the dipole and its orientation. At the origin, the contributions from the positive and negative charges of the dipole can be combined to determine the net electric field, typically resulting in a field that varies with the angle relative to the dipole axis and diminishes with distance.
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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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MatinSAR
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Then it is equal to ##Pr^{-3}##, Am I right?!kuruman said:
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Show me the math.
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MatinSAR
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kuruman said:
Sorry that ##r^{-3}## was related to another part.
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MatinSAR
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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##
##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##
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Yes. What is your final answer when you put it all together?MatinSAR said:
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MatinSAR
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##\vec P## , I guess.kuruman said:
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Sorry, not that. I meant putting together the final expression ##\vec E=-\vec{\nabla}\psi=?##
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MatinSAR
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I am trying to solve ... I will send the work.kuruman said:
Thanks again for your help Prof.Kuruman
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MatinSAR
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That's it. Good job!
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MatinSAR
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Thanks a lot! Have a good day.kuruman said:
FAQ: Find the Electric field due to an electric dipole at the origin
What is an electric dipole?
An electric dipole consists of two equal and opposite charges separated by a small distance. It is characterized by a dipole moment, which is a vector quantity pointing from the negative charge to the positive charge.
How do you calculate the electric field due to an electric dipole at a point on the axial line?
The electric field due to a dipole at a point on the axial line (along the line connecting the two charges) is given by \( E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{r^3} \), where \( p \) is the dipole moment and \( r \) is the distance from the dipole to the point where the field is being calculated.
How do you calculate the electric field due to an electric dipole at a point on the equatorial line?
The electric field due to a dipole at a point on the equatorial line (perpendicular bisector of the line connecting the two charges) is given by \( E = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{r^3} \), where \( p \) is the dipole moment and \( r \) is the distance from the dipole to the point where the field is being calculated.
What is the direction of the electric field due to an electric dipole at the origin?
The direction of the electric field due to an electric dipole at the origin depends on the position of the point where the field is being calculated. Along the axial line, the field points away from the dipole if the point is on the positive charge side, and towards the dipole if the point is on the negative charge side. Along the equatorial line, the field is perpendicular to the dipole moment and points away from the dipole axis.
How does the distance from the dipole affect the electric field strength?
The electric field strength due to an electric dipole decreases with the cube of the distance from the dipole. Specifically, the field strength is inversely proportional to \( r^3 \), where \( r \) is the distance from the dipole to the point where the field is being calculated.