Question
Given a uniformly charged plane/sheet of surface charge density, $\sigma=2\times10^{17}\text{C/m}^2.$Image
  1. Find the electric field intensity at a point A, 5mm away from the sheet on the left side.
  2. Given a straight line with three points X, Y and Z placed 50cm away from the charged sheet on the right side. At which of these points, the field due to the sheet remain the same as that of point A and why?

Answer

  1. At A, $\text{E}=\frac{\sigma}{2\epsilon_0}=\frac{2\times10^{17}\text{Cm}^{-2}}{2\times8.854\times10^{-12}\text{C}^2\text{N}^{-1}\text{m}^{-2}}$
$\text{E}=1.1\times1028 \text{N/C}$
Directed away from the sheet.
  1. Point Y, Because at 50cm, the charge sheet acts as a finite sheet and thus the magnitude remains same towards the middle region of the planar sheet.

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