- #1

gj2

## 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##).

## Homework Equations

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

## 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.

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