In the figure, a very large plane sheet of positive charge is shown. $$P_1$$ and $$P_2$$ are two points at distance $$l$$ and $$2l$$ from the charge distribution. If $$\sigma$$ is the surface charge density, then the magnitude of electric fields $$E_1$$ and $$E_2$$ at $$P_1$$ and $$P_2$$ respectively are

For a large plane sheet of positive charge with a uniform surface charge density $\sigma$, the electric field intensity at nearby points is determined using Gauss's Law.

1. Electric Field Expression

The electric field E generated by an infinite non-conducting plane sheet is given by the formula:

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

Where:

• $$\sigma$$ = Surface charge density
• $$\varepsilon_0$$ = Permittivity of free space

2. Independence of Distance

A key characteristic of a very large (infinite) plane sheet is that the electric field it produces is uniform. This means the magnitude of the field does not depend on the distance r from the sheet, provided the distance is small compared to the dimensions of the sheet.

3. Comparison of Points $$P_1$$ and $$P_2$$

Since points $$P_1$$ and $$P_2$$ are at different distances but are both near the large sheet:

• The field at $$P_1$$ is $$E_1 = \frac{\sigma}{2\varepsilon_0}$$
• The field at $$P_2$$ is $$E_2 = \frac{\sigma}{2\varepsilon_0}$$

Final Conclusion:

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

Correct Option: (C)