The electric field intensity at point near and outside the surface of a charged conductor of any shape is `E_(1)` the electric field intensity due to uniformly charged infinite thin plane sheet is `E_(2)` the relation between `E_(1)` and `E_(2)` is

Options

A. `2E_(1)=E_(2)`
B. `E_(1)=E_(2)`
C. `E_(1)=2E_(2)`
D. `E_(1)=4E_(2)`

Answer

C

Explanation

Electric field intensity at a point mear and outside the surface of any charged conductor is given by
`E_(1)=(sigma)/(epsilon_(0))" …(i)"`
where, `sigma` is charge density (charge per unit area) at a point.
Electric field intensity due to uniformaly charged infinite thin plane sheet is given by
`E_(2)=(sigma)/(2epsilon_(0))" ...(ii)"`
From Eqs. (i) and (ii), we have
`E_(1)=(2)/(2)xx(sigma)/(epsilon_(0))=2xxE_(2)`
`rArr" "E_(1)=2E_(2)`