Potential of infinite sheet with thickness

[susdu]

Homework Statement

Describe the potential inside and outside an infinite insulating sheet with uniform density ρ and thickness d, as a function of x (distance from the center of the sheet). zero potential has been set at its center. What is the potential on the surface of the sheet?

Homework Equations

Potential and E.field definitions.

The Attempt at a Solution

I know that inside the sheet (x<\frac{d}{2}) the field is given by

E=\frac{ρx}{ε_0}.

so, potential inside the sheet (with $V_i=0$ at $x=0$) is:

$V_f=-\int^x_0 E\,ds=-\frac{ρx^2}{ε_0}$

similarly, outside the sheet (x>\frac{d}{2}) field is

E=\frac{ρd}{ε_0}.

however, I'm not sure about the expression for outside the sheet, is it

$V_f=-\int^\frac{d}{2}_0 E\,ds-\int^x_\frac{d}{2} E\,ds$

?

 

[Gloyn]

I would use the Gauss Law. Let's consider a cylindrical surface with it's axis perpendicular to the sheet that "sticks out" of the sheet equally. Gauss Law states, that:

\epsilon _0 E \int ds = Q

And Q=density of charge x volume.
From symmetry we know that there's no flux through the curved surface of the cyllinder, so we have flux only through bases of the cyllinder:

\epsilon _0 E 2\pi R^2 = \rho \pi R^2 d

So E=\frac{\rho d}{2\epsilon _0}

And E is constant everywhere outside the sheet.

 

[susdu]

I already know the expressions for E.field outside/inside the sheet.
I need an expression for potential.

 

[haruspex]

$V_f=-\int^\frac{d}{2}_0 E\,ds-\int^x_\frac{d}{2} E\,ds$?

Looks right to me.