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Electric field of a thick infinite non-conducting plate

gj2
1. Homework Statement
An infinite non-conducting plate of thickness ##d## lies in the ##xy## plane. The bottom surface of the plate lies in the plane ##z=0##. The charge density of the plate is ##\rho=\rho_0 z /d~,~\rho_0>0##.
Find the electric field in the regions ##z<0~~,~~0<z<d~~,~~z>d## and the potential (assume it's zero at ##z=0##).

2. Homework Equations
Poisson's equation ##{\displaystyle {\nabla }^{2}\varphi =-{\frac {\rho }{\varepsilon_0 }}}## and ##\vec{E}=-\operatorname{grad} \varphi##

3. The Attempt at a Solution
The charge distribution does not depend on ##x## or ##y## and the plate is symmetric therefore both the potential and the electric field can't depend on ##x## or ##y##. We know that ##\vec{E}=-\operatorname{grad} \varphi##. The Poisson's equation for the region inside the plate is
$$\frac{\partial^2 \varphi}{\partial z^2}=-\frac{\rho_0 z}{\varepsilon_0 d}$$
Assuming ##\varphi(z=0)=0## we obtain
$$\varphi(z)=C_1z-\frac{\rho_0 z^3}{6\varepsilon_0 d}~~~,~~~0<z<d$$
Poisson's equation for the region outside the plate
$$\frac{\partial^2 \varphi}{\partial z^2}=0$$
Therefore
$$\varphi(z)=C_2z$$
The potential must be continuous everywhere and particularly at ##z=d## so
$$C_2=C_1-\frac{\rho_0 d}{6\varepsilon_0}$$
In other words
$$\varphi(z)=\left\{\begin{matrix}
Cz-\frac{\rho_0 z^3}{6\varepsilon_0 d} & z\in(0,d)\\
Cz-\frac{\rho_0 dz}{6\varepsilon_0} & z\notin(0,d)
\end{matrix}\right.$$
But I can't figure out what other condition should I impose in order to obtain the last missing constant.
 
Last edited by a moderator:

Answers and Replies

ehild
Homework Helper
15,361
1,778
1. Homework Statement
An infinite non-conducting plate of thickness ##d## lies in the ##xy## plane. The bottom surface of the plate lies in the plane ##z=0##. The charge density of the plate is ##\rho=\rho_0 z /d~,~\rho_0>0##.
Find the electric field in the regions ##z<0~~,~~0<z<d~~,~~z>d## and the potential (assume it's zero at ##z=0##).

2. Homework Equations
Poisson's equation ##{\displaystyle {\nabla }^{2}\varphi =-{\frac {\rho }{\varepsilon_0 }}}## and ##\vec{E}=-\operatorname{grad} \varphi##

3. The Attempt at a Solution
The charge distribution does not depend on ##x## or ##y## and the plate is symmetric therefore both the potential and the electric field can't depend on ##x## or ##y##. We know that ##\vec{E}=-\operatorname{grad} \varphi##. The Poisson's equation for the region inside the plate is
$$\frac{\partial^2 \varphi}{\partial z^2}=-\frac{\rho_0 z}{\varepsilon_0 d}$$
Assuming ##\varphi(z=0)=0## we obtain
$$\varphi(z)=C_1z-\frac{\rho_0 z^3}{6\varepsilon_0 d}~~~,~~~0<z<d$$
Poisson's equation for the region outside the plate
$$\frac{\partial^2 \varphi}{\partial z^2}=0$$
Therefore
$$\varphi(z)=C_2z$$
The potential must be continuous everywhere and particularly at ##z=d## so
$$C_2=C_1-\frac{\rho_0 d}{6\varepsilon_0}$$
In other words
$$\varphi(z)=\left\{\begin{matrix}
Cz-\frac{\rho_0 z^3}{6\varepsilon_0 d} & z\in(0,d)\\
Cz-\frac{\rho_0 dz}{6\varepsilon_0} & z\notin(0,d)
\end{matrix}\right.$$
But I can't figure out what other condition should I impose in order to obtain the last missing constant.
For a second-order equation, you need to give two boundary conditions. What can they be?
Do not forget that the Poisson equation is derived from the original Maxwell equations, so you can use Gauss' Law.
 
TSny
Homework Helper
Gold Member
12,153
2,686
If you are going to find ##\varphi## by integrating Poisson's equation, you have three separate regions to deal with. You cannot assume that the mathematical expression for ##\varphi(z)## when ##z>d## is the same as for when ##z<0##.

Adding a bit to ehild's hint, it might be easier to first find ##\mathbf{E}## and then use it to find ##\varphi(z)##.
 
gj2
For a second-order equation, you need to give two boundary conditions. What can they be?
Do not forget that the Poisson equation is derived from the original Maxwell equations, so you can use Gauss' Law.
If you are going to find ##\varphi## by integrating Poisson's equation, you have three separate regions to deal with. You cannot assume that the mathematical expression for ##\varphi(z)## when ##z>d## is the same as for when ##z<0##.

Adding a bit to ehild's hint, it might be easier to first find ##\mathbf{E}## and then use it to find ##\varphi(z)##.
Yes, thank you both. I realized beforehand that I did a mistake. In fact, it is much more complicated. But anyways I managed to solve it. In order to obtain the constants I used three things: 1) the fact that the electric field outside the plate is symmetrical w.r.t the plate (and not just constant) 2) Gauss law where the two bases of the Gaussian cylinder/box are outside the plate 3) Gauss law where one base is inside the plate and the other base outside it.
 

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