Charge density required to create an electric field

In summary, the conversation discusses finding the charge density necessary to create a given electric field. The relevant equation for this is ##\rho =\varepsilon_0\, \nabla \cdot \vec E##, which results in the expression ##\rho(x,y,z)=\epsilon_0(2ax+c)## for the given electric field. The meaning of this result is that in order to create the given electric field, one would need to place linearly increasing amounts of charges in space as the distance from the ##yz##-plane increases, with the right sign for the charges.
  • #1
Cepterus
29
0

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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  • #2
You will need a different relevant equation: something like ##\rho =\varepsilon_0\, \nabla \cdot \vec E##
 
  • #3
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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