The electric potential at surface of thin non-conducting sheet with charge density `sigma` is `V_(0)`. Show that the electric potential at a distance x from infinite sheet can be written as
`V = V_(0) - (sigma)/(2 in_(0)) x`

Electric field due to an infinite non-conducting sheet of surface charge density sigma , at a distance r from it is

A large metallic plate is facing a charged sheet having charge density sigma placed parallel to plate at a distance l from the plate.Potential at point P at a distance x from the sheet is

Two infinite parallel,non- conducting sheets carry equal positive charge density sigma ,One is placed on the other at distance x=a Take potential V=0 at x=0 then

An infinite conducting sheet has surface charge density sigma . The distance between two points is r. The potential difference (V_A - V_B) between these point is

A large metallic plate is facing a charged sheet having charge density "sigma" placed parallel to plate at a distance "l" from the plate.Potential at "point "P" at a distance [x" from the sheet is "]

There is an infinite non conducting sheet of charge having uniform charge density sigma . The electric field at a point P at a distance x from the sheet is E_(0) . Point O is the foot of the perpendicular drawn from point P on the sheet. A circular portio of radius r lt lt x centered at O is removed from the sheet. Now the field at point P becomes E_(0)-DeltaE . Find DeltaE .

The electric potential at 0.9m from a point charge is +50 V . What is the magnitude and sign of the charge ?

A conductor with a cavity is charged positively and its surface charge density is sigma . If E and V represent the electric field and potential, then inside the cavity