If q is the charge per unit area on the surface of a conductor, then the electric field intensity at a point on the surface is

Options

A. `(q/(epsilon_0))` normal to surface
B. `(q/(2epsilon_0))` normal to surface
C. `(q/(epsilon_0))` tangential to surface
D. `(q/(2epsilon_0))` tangential to surface

Answer

A

Explanation

For conductor
`E= sigma/(epsilon_0) ` [ `sigma rarr` surface charge density]
Here `sigma = q =` charge per unit area
`:. E = q/(epsilon_0)` [normal to the surface].