State Gauss's theoram in electric field at a point due to a uniformaly charged infinite plane thin sheet.

Explanation

For Gauss. theorem , see point number 47 under the heading "Chapter At A Glance" Consider an infinite thin plane sheet having a surface density ` sigma. ` To find electric field at a point P situated at a normal distance r from the sheet , consider an imaginary cylinder of cross-section ara ds around point P and length 2r, passing through the sheet ,as the Gaussian surface.
From symmetry consideration , only side faces1 and 2 of cylinder contribute towards the flux because here
` oversetto E and hatn ` parallel but the curved surface of cylinder does not contributed towards the flux because here `oversetto E and hatn ` are mutually perpendicular.
` therefore ` Total electric flux ` phi_in =2 E ds`
As per Gauss theorem total electric flux `pi_in =(1)/( in_0) ` (charge encloesd) `=(1)/( in _0) .(sigma ds) `
Comparing (i) and (ii) , we get
` 2 E ds =(sigma)/( in_0) . ds rArr E= ( sigma)/( 2 in _0) `
Thus, the electric field at a point due to a uniformaly charged infinite plane sheet is independent of the distance from it .
Vectorially ` oversetto (E) =(sigma)/(2 in _0) hatr `
Thus for (i) positively charged sheet electric field `oversetto E ` is directed outwards and for (ii) negatively charged sheet the field is directed inwards.