Charge density required to create an electric field

AI Thread Summary
To find the charge density required to create the given electric field, the relationship between charge density and electric field is established using the equation ρ = ε₀ ∇ · E. The calculated charge density is ρ(x,y,z) = ε₀(2ax + c). This indicates that the charge density varies linearly with respect to the x-coordinate, suggesting that more charge is needed as one moves away from the yz-plane. The discussion emphasizes the need to understand how charge distribution influences the electric field configuration. Understanding this relationship is crucial for solving problems related to electric fields and charge distributions.
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Homework Statement


Given an electric field $$\vec E(x,y,z)=\begin{pmatrix}ax^2+bz\\cy\\bx\end{pmatrix},$$with nonzero constants ##a,b,c##, I am supposed to find the charge density ##\rho(x,y,z)## which is necessary to create this field ##\vec E##.

Homework Equations


##\rho=\frac{\mathrm dq}{\mathrm dV}##

The Attempt at a Solution


I know that electric fields are created by charges, but I don't understand the connection between charge density and an electric field. As far as I understand the above formula, charge density is simply the amount of charge in a certain volume. How does this relate to electric fields?
 
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You will need a different relevant equation: something like ##\rho =\varepsilon_0\, \nabla \cdot \vec E##
 
BvU said:
You will need a different relevant equation: something like ##\rho =\varepsilon_0\, \nabla \cdot \vec E##
Thanks! I get the result ##\rho(x,y,z)=\epsilon_0(2ax+c)##. If this is correct, what does it actually mean? Like, to create ##\vec E##, should I start with zero charge on the ##yz##-plane and then place linearly more and more charges in space the larger my distance to the ##yz##-plane gets? (with the right sign, of course)
 
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