Home
Class 12
PHYSICS
If q is the charge per unit area on the ...

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

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

Text Solution

Verified by Experts

The correct Answer is:
A
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].
Promotional Banner

Similar Questions

Explore conceptually related problems

The charge deposited per unit area of the surface is called

If an isolated infinite plate contains a charge Q_1 on one of its surfaces and a charge Q_2 on its other surface, then prove that electric field intensity at a point in front of the plate will be Q//2Aepsilon_0 , where Q = Q_1+Q_2

If an isollated infinite sheet contains charge Q_(1) on its one surface and charge Q_(2) on its other surface, then prove that electric field intensity at a point in front of sheet will be Q/(2A epsi_(0)) , where Q=Q_(1)+Q_(2)

The electric field intensity on the surface of a charged conductor is

A closed conducting sphere is positively charged such that the surface charge density is 17.7xx10^(-12)C//m^(2) . If the dielectric constant of the medium surrounding the conductor is 100, then electric field intensity at a point just outside it will be

Why do we always get electric field intensity near a charged metallic surface of magnitude twice of the electric field intensity near a thin sheet of charge with same surface charge density as of metal plate?