A charge is uniformly distributed over a large iron plate. The electric field at a point close to the centre of the plate is 40 V m -1. If the iron plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge, then the electric field at the same point will be

Question

A charge is uniformly distributed over a large iron plate. The electric field at a point close to the centre of the plate is 40 V m -1. If the iron plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge, then the electric field at the same point will be (A) 1 V m -1 (B) 10 V m -1 (C) 40 V m -1 (D) 80 V m -1

Tardigrade Admin · Accepted Answer

E=(σ/ϵ0) (for metal plate) σ is the surface charge density. It does not depend on the type of conductor.

$1 V m^{- 1}$

$10 V m^{- 1}$

$40 V m^{- 1}$

$80 V m^{- 1}$

$E = \frac{σ}{ϵ _{0}}$ (for metal plate)
$σ$ is the surface charge density. It does not depend on the type of conductor.

1Vm−1

10Vm−1

40Vm−1

80Vm−1

E=ϵ0​σ​ (for metal plate) σ is the surface charge density. It does not depend on the type of conductor.

$1 V m^{- 1}$

$10 V m^{- 1}$

$40 V m^{- 1}$

$80 V m^{- 1}$

$E = \frac{σ}{ϵ _{0}}$ (for metal plate)
$σ$ is the surface charge density. It does not depend on the type of conductor.