Question

1. The surface density of charge on a conductor situated in air, is 2x10^-4 C/
m2
The electric intensity at a point very near to its surface is *​

Answer 1

electric field at a point near to its surface is 2.258 × 10^7 N/C.

It has given that surface charge density of a conductor situated in air , σ = 2 × 10¯⁴ C/m²

we have to find the electric field intensity at a point near to its surface.

surface charge density is the rate of change of charge with respect to area.

⇒dQ/dA = σ

⇒dQ = σdA

now from Gauss law, electric flux , Φ = Q/∈_0

⇒E.dA = σdA/∈_0

⇒ E = σ/∈_0

here, ∈_0 is permittivity of medium (I.e.,air ) i.e., 8.854 × 10¯¹² C²/Nm²

now, E = (2 × 10¯⁴ C/m²)/(8.854 × 10¯¹² C²/Nm²)

= (0.2258) × 10^8 N/C

= 2.258 × 10^7 N/C

therefore, electric field at a point near to its surface is 2.258 × 10^7 N/C.

Answer 2

Answer:

electric field at a point near to its surface is 2.258 × 10^7 N/C.

It has given that surface charge density of a conductor situated in air , σ = 2 × 10¯⁴ C/m²

we have to find the electric field intensity at a point near to its surface.

surface charge density is the rate of change of charge with respect to area.

⇒dQ/dA = σ

⇒dQ = σdA

now from Gauss law, electric flux , Φ = Q/∈_0

⇒E.dA = σdA/∈_0

⇒ E = σ/∈_0

here, ∈_0 is permittivity of medium (I.e.,air ) i.e., 8.854 × 10¯¹² C²/Nm²

now, E = (2 × 10¯⁴ C/m²)/(8.854 × 10¯¹² C²/Nm²)

= (0.2258) × 10^8 N/C

= 2.258 × 10^7 N/C

therefore, electric field at a point near to its surface is 2.258 × 10^7 N/C.