Can \(\vec{E} = E_0 \cdot (-y,x, z)\) Be an Electrostatic Field?

[henrybrent]

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I have a question which is:Let \vec{E} = E_0 \cdot (-y,x, z) Can \vec{E} be an electrostatic field? if yes, find the charge density which generates this field. If not, find the magnetic field which generates itand,

Let \vec{E} = E_0 \cdot \vec{r} ) Can \vec{E} be an electrostatic field? if yes, find the charge density which generates this field. If not, find the magnetic field which generates itI have no idea where to start, any help is appreciated.

 

[ShayanJ]

An electrostatic field, is an electric field for which we can find an scalar field(a function of spatial coordinates)\phi such that \vec E=-\vec \nabla \phi. Now if I take the curl of this equation, I get \vec \nabla \times \vec E=0(because the curl of the gradient of a scalar field is always zero). So you should see whether the curl of the given electric fields are zero or not.

 

[henrybrent]

I am not sure how to take curl of the electric fields, sorry.

I am not sure what E_0 denotes? is that merely a constant?

 

[ShayanJ]

<br /> \vec \nabla \times \vec E=\vec \nabla \times (E_x,E_y,E_z)=(\frac{\partial E_z}{\partial x}-\frac{\partial E_y}{\partial z})\hat x+(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x})\hat y+(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y})\hat z<br />
And E_0 is only a constant.