Find the electric field intensity due to infinite uniformly charged non-conducting sheet at a point near it, using Gauss’s law. Explain the dependence of electric field intensity.
Answers
Answer:
To find the electric field due to infinite uniform charged plane sheet we take a cylindrical Gaussian surface such that the surface passes perpendicular to the plane of the sheet. The surface charge density of the sheet is taken as σ . The electric field is normally outward to the plane sheet and is of same magnitude in both the directions. The cross sectional area of the Gaussian surface is taken to be A. The Gaussian cylinder consists of three surfaces P, S2 and P’ but the electric flux because of the sheet linked with the surface S2 is 0.
Hence, the total flux linked through the Gaussian surface is :
ΦT = electric flux through S2 + electric flux through P’ + electric flux through P
ΦT = 0 + EA cos 0° + EA cos 0°
ΦT = 2 EA …….(i)
The gauss theorem states that:
Where,
‘q’ is the charge is the charge on the plate given as, q = σA
ϕ is the flux through the Gaussian surface,
is the permittivity in air.
⇒
……..(ii)
Equating equation (i) and (ii) we get,
This is the electric field due to uniformly charged infinite plane sheet.
The direction of field is perpendicular to the plane of the sheet and points outward in the case of positive charge density. Whereas, in the case of negative charge it is in the opposite direction .i.e. perpendicularly inward.