Question

Two large charged plane sheets of charge
densities sigma and -2 sigma C/m² are arranged
vertically with a separation of d between
them. Deduce expressions for the electric
field at points (i) to the left of the first
sheet, (ii) to the right of the second sheet
and (iii) between the two sheets.​

Answer 1

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Answer 2

Given :

Two charged plane sheets of charge densities,

$\sigma \: and \: - 2 \sigma$

To find :

Electric field to left of first sheet, right of second sheet and between both sheets.

Solution :

It is given that there are two sheets,

both have different charge densities,

$\sigma \: and \: - 2 \sigma$

We can see the arrangement of both the sheets as given in figure.

We know that Electric field due to a conducting sheet is:

$E = \frac{ \sigma \: }{ 2\epsilon _{0}}$

(equation 1)

In this case there are two sheets with two different charge densities.

So, according to Superposition principal :

$\bf \: E = E _{1} + E_{2}$

(equation 2)

The resultant electric field is vector sum of all the electric fields in a system.

At left of first sheet,

Electric field by left sheet is away from it and

Electric field by right sheet is towards its direction.

At right side of second sheet,

Electeic field by right sheet is towards it and

Electric field by left sheet is away from its direction.

At region between both sheets,

Electric field by both sheets is towards second sheet.

By equation 1 and equation 2,

Electric field at left side of first sheet :

$E = - \frac{ \sigma \: }{ 2\epsilon _{0}} \hat r \: + \frac{ 2\sigma \: }{ 2\epsilon _{0}} \hat r$

$E = \frac{ \sigma \: }{ 2\epsilon _{0}} \hat r$

Electric field at right side of second sheet :

$E = - \frac{ \sigma \: }{ 2\epsilon _{0}} \hat r \: - \frac{ 2\sigma \: }{ 2\epsilon _{0}} \hat r$

$E = - \frac{ 3\sigma \: }{ 2\epsilon _{0}} \hat r$

Electric field between both sheets :

$E = - \frac{ \sigma \: }{ 2\epsilon _{0}} \hat r \: + \frac{ 2\sigma \: }{ 2\epsilon _{0}} \hat r$

$E = \frac{ \sigma \: }{ 2\epsilon _{0}} \hat r$