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**Hint:**In uniform conducting sheet the charge goes on the surface of the sheet and for non-conducting sheet the charge will remain on its position. Here consider the thickness of the sheet to be approximately zero that means it will have only area and no volume. (Charge is spread on the sheet).

**Complete step by step answer:**

Find out the formula for electric field due to a non-conducting infinite plane having a uniform charge density:

Here the flux would be:

$\phi = 2E \times 2\pi rl$;

Here:

E = Electric Field;

r = distance;

l = length;

We know that electric flux is equal to the sum of all charges divided by the free space:

$\dfrac{Q}{{{\varepsilon _o}}} = 2E \times 2\pi rl$;

We know that the surface charge density is charge upon area:

$\sigma = \dfrac{Q}{A}$ ;

$\sigma \times A = Q$;

Put the above relation in the formula for flux$\dfrac{Q}{{{\varepsilon _o}}} = 2E \times 2\pi rl$:

$\dfrac{{\sigma \times A}}{{{\varepsilon _o}}} = 2E \times 2\pi rl$;

Put the value in the above equation:

$\dfrac{{\sigma \times A}}{{{\varepsilon _o}}} = 2E \times 2\pi rl$;

$ \Rightarrow \dfrac{{\sigma \times 2\pi rl}}{{{\varepsilon _o}}} = 2E \times 2\pi rl$;

Cancel out the common in the above equation:

$ \Rightarrow \dfrac{\sigma }{{{\varepsilon _o}}} = 2E$;

$ \Rightarrow E = \dfrac{\sigma }{{2{\varepsilon _o}}}$;

The above equation is the equation of the electric field due to a non-conducting infinite plane having a uniform charge density. Here, we can see that the electric field has no relation with the distance “r”.

**Hence, the correct answer is option (B).**

**Note:**Here we need to first find out the formula for electric flux through a given area and then put the found-out flux in relation with electric field and solve for the Electric field due to infinite plane non conducting sheet.

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