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 \(σ \) is the surface charge density, then the magnitude of electric fields \(E_1\) and \(E_2\) and \(P_1\) and \(P_2\) respectively are
To solve this problem, we need to determine the electric field due to a very large plane sheet of positive charge at two points, \( P_1 \) and \( P_2 \), located at distances \( l \) and \( 2l \) respectively from the charge distribution.
Concept
The electric field due to an infinitely large plane sheet of charge with surface charge density \( \sigma \) is given by:
E = \frac{\sigma}{2\varepsilon_0}
where \( \varepsilon_0 \) is the permittivity of free space.
Explanation
The electric field due to a plane sheet of charge is uniform and does not depend on the distance from the sheet. This is because of the symmetry and infinite extent of the plane.
Thus, both points \( P_1 \) and \( P_2 \) will experience the same electric field E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}.
Conclusion
Therefore, the magnitude of the electric fields at P_1 and P_2 are both equal to \frac{\sigma}{2\varepsilon_0}.
The correct answer is: E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}.
