1. Electric field due to a very large plane sheet of charge

For an infinite (or very large) plane sheet with uniform surface charge density $\sigma$, the electric field on either side is constant and is given by

$E = \frac{\sigma}{2\varepsilon_0}$

This result is obtained using Gauss's law.

2. Using Gauss's law briefly

Choose a cylindrical Gaussian surface (pillbox) of cross-sectional area $A$ passing through the sheet.

• Total flux through the two flat faces: $\Phi = EA + EA = 2EA$
• Charge enclosed: $q_{\text{enc}} = \sigma A$

By Gauss's law,

$2EA = \frac{\sigma A}{\varepsilon_0}$

So,

$E = \frac{\sigma}{2\varepsilon_0}$

3. Apply at points $P_1$ and $P_2$

Since the field of an infinite plane sheet does not depend on distance from the sheet,

• at $P_1$ (distance $l$): $E_1 = \frac{\sigma}{2\varepsilon_0}$
• at $P_2$ (distance $2l$): $E_2 = \frac{\sigma}{2\varepsilon_0}$

Thus,

$E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}$

4. Check options

• A: Incorrect
• B: Incorrect
• C: Correct
• D: Incorrect

Therefore, the correct option is C.