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## Main Question or Discussion Point

I started having this doubt since i saw the derivation of potential across a capacitor using this method. So.. here i ask:

Let there be two large sheets made of insulators, and charged to a planar charge density, [itex]\sigma[/itex] (Sheet 1) and [itex]2\sigma[/itex] (Sheet 2). Since they are insulators, when placed in proximity to each other, it can be assumed that the charges will not redistribute on either of them.

Now, we put them one against the other with a separation 'd' as is done for a capacitor. Now, i take a cylindrical surface perpendicular to Sheet 1, such that one end of the cylinder is within the thickness of the sheet, with radius 'r'. Then, the flux through the cylinder is only through the other surface. Hence,

[tex]

\Phi = E(4\pi r^2)

[/tex]

Using Gauss' law:

[tex]

E(4 \pi r^2) = \frac{q}{\epsilon_o}

[/tex]

and hence,

[tex]

E = \frac{\sigma}{\epsilon_o}

[/tex]

To calculate the potential, i use [itex]dV = E.dl[/tex]

[tex]

V = \frac{\sigma d}{\epsilon_o}

[/tex]

However, if i do the same thing but by making the cylindrical surface on the other plate, i'm getting:

[tex]

V = \frac{2\sigma}{\epsilon_o}

[/tex]

How am i getting two potential differences for the same system? What am i doing wrong?

Let there be two large sheets made of insulators, and charged to a planar charge density, [itex]\sigma[/itex] (Sheet 1) and [itex]2\sigma[/itex] (Sheet 2). Since they are insulators, when placed in proximity to each other, it can be assumed that the charges will not redistribute on either of them.

Now, we put them one against the other with a separation 'd' as is done for a capacitor. Now, i take a cylindrical surface perpendicular to Sheet 1, such that one end of the cylinder is within the thickness of the sheet, with radius 'r'. Then, the flux through the cylinder is only through the other surface. Hence,

[tex]

\Phi = E(4\pi r^2)

[/tex]

Using Gauss' law:

[tex]

E(4 \pi r^2) = \frac{q}{\epsilon_o}

[/tex]

and hence,

[tex]

E = \frac{\sigma}{\epsilon_o}

[/tex]

To calculate the potential, i use [itex]dV = E.dl[/tex]

[tex]

V = \frac{\sigma d}{\epsilon_o}

[/tex]

However, if i do the same thing but by making the cylindrical surface on the other plate, i'm getting:

[tex]

V = \frac{2\sigma}{\epsilon_o}

[/tex]

How am i getting two potential differences for the same system? What am i doing wrong?