Electric Field and Electric Dipole Moment for a Dipole

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Philosophaie
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The Magnetic Dipole Moment for a Magnetic Field for a dipole oriented on the x-y axis is:
##\bar m = |m| \hat z##
The Magnetic Field is:
##\bar B = \frac{\mhu_0}{4 * \pi * |\bar r|^5} * 3 * \bar r * (\bar m . \bar r) - \bar m * |\bar r|^2##
Vector Potential is:
##\bar A = \frac{\mhu_0}{4 * \pi * |\bar r|^3} * (\bar m X \bar r)##
How do you find the Electric Field and Electric Dipole Moment, ##\bar p## for the above dipole?

This is not a homework problem.
 
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Since this is obviously a static problem you have to solve the electrostatics part independently from the magnetic field since in the static case the Maxwell equations decouple in those for the electric an magnetic components.
 
How do you solve for a static scalar potential using the Magnetic Dipole Moment to then solve for the Electric Field?
 
I don't understand, what you mean. To solve for the electric field you need the charge distribution. It's unaffected by a static magnetic dipole moment.
 
Is the Electric Field not a function of the scalar and vector potential?
 
Calculating the 4-scalar potential, ##\phi##, is my dilemma for a dipole.

When I find the equation of 4-Scalar Potential for a dipole I can can calculate the 4-Vector Potential, ##A(\phi/c,\bar r)##.

I just need help formulating the Equation for 4-##\phi##
 
An electrostatic dipole usually is describe by a scalar potential (in Heaviside-Lorentz units)
$$\phi(\vec{x})=\frac{\vec{p} \cdot \vec{x}}{4 \pi |\vec{x}|^3},$$
and a magnetic one by a vector potential,
$$\vec{A}(\vec{x})=\frac{\vec{m} \times \vec{x}}{4 \pi |\vec{x}|^3},$$
where ##\vec{p}## is the electric and ##\vec{m}## the magnetic dipole moment of the charges, currents, and permanent magnets.
 
How do you define the Electrostatic Dipole Moment ##\bar p##. The magnitude for a Dipole would be ##|\bar p| = Q*|\bar d|## where d is the separation of poles of the positive and negative Dipole terminals. Is the direction of the Electrostatic Dipole Moment ##\hat p:\hat x=x-d/2, \hat y=0## and ##\hat z=0##?
 
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vanhees71 said:
##\vec##p=∫R3d3##\vec##x ρ(##\vec##x)\##\vec##x.
##\int_{\mathbb{R}^3} \mathrm{d}^3\vec x = 1 \hat x## ##0 \hat y## ##0 \hat z##
##\frac{d \rho}{dt}=-grad.\bar J##
##\vec{x}=(x-(d/2)) \hat x## ##0 \hat y## ##0 \hat z##
 
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Philosophaie said:
Where am I going wrong?
I cannot read what you wrote. It's somehow garbled. I don't know, what went wrong. Are you, maybe, using the App on a mobile device?