Question

Use gauss law to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities +signa and -sigma respectively

Answer 1

Given :

No of uniformly charged large parallel sheets = 2

Surface charge densities of the two sheets = $+\sigma$  and $-\sigma$

To Find :

The expression for the electric field between  these two sheets = ?

Solution :

Consider a plane sheet with surface density $+\sigma$ ( diagram is shown in the attached fig )  , we have to find electric field at point P .

Choose a Gaussian surface for this as follows :

Draw a plane surface A passing through P and parallel to charge  sheet .

Draw a cylinder with this surface as cross section and equally extend it to other side .

The electric field at P will act outwards and according to Gauss Law  the flux through A is :

$\phi= \vec E .\Delta \vec S$      ( E = electric field  and $\Delta \vec S$ = cross sectional area of A and A')

   =$E\Delta S$

Total flux :

$\phi = \int\int\vec E .\vec {ds}$

  =$E.\Delta S + E . \Delta S+ 0 =2 E .\Delta S$

(∴at the point of curved surface , the field and outward normal are at 90° , hence $\vec E .\Delta \vec S = 0$ )

We also have :

$E = \frac{q}{\epsilon_0}$  $= \frac{\sigma . \Delta S}{\epsilon_0}$

So, $2E.\Delta S = \frac{\sigma \Delta S}{\epsilon_0}$

Or, $E = \frac{\sigma }{2 \epsilon_0}$

Similarly for the negative charged sheet  :

$E = \frac{-\sigma }{2 \epsilon_0}$  ( in the inward direction )

So, total $\vec E = \frac{\sigma}{2 \epsilon_0 } - ( \frac{-\sigma}{2 \epsilon_0 } )$

                   = $\frac{\sigma}{ \epsilon_0 }$

So, the expression for the electric field between these two given uniformly  charged parallel sheets is $\vec E = \frac{\sigma}{ \epsilon_0 }$ .

Answer 2

Answer:

this is the answer

Explanation:

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