Class 12 Physics, CBSE and UP Board, Electrostatic Potential and Capacitance, Additional Exercise,Chapter-2 Exercise 2.16,NCERT Solutions for Class 12 Physics
(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.]
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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.