(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by ( E₂ -E₁ ). ñ =σ/ ε0 where ñ is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of ñ is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ nˆ /ε0. (b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

Electric field on one side of a charged body is E₁ and electric field on the other side of the same body is E2. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

Ē₁= (σ /2 ε₀ )ñ ——————–Eq-1

Where ,

ñ = Unit vector normal to the surface at a point

σ= Surface charge density at that point

Electric field due to the other surface of the charge body ,

Ē₂= – (σ /2 ε₀ )ñ ——————–Eq-2

Electric field at any point due to the two surfaces,

Ē₂ –  Ē₁= (σ /2 ε₀) ñ + (σ /2 ε₀ )ñ   =  (σ /ε₀ )ñ

(Ē₂ –  Ē₁) ñ = σ /ε₀ ———————-Eq-3

Since inside a closed conductor ,Ē₁=0.

Therefore  Ē=Ē₂= (σ /2 ε₀)ñ 

,the electric field just outside the conductor is σ /ε₀ ñ

Ans (b).

When a charged particle is moved from one point to the other on a closed loop ,the work done by the electrostatic field is zero. Hence ,the tangential component of electrostatic field is continuous from one side of a charged surface to the other.