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(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by (E_2-E_1) hat n= (sigma)/(epsi_0) where hatn is a unit vector normal to the surface at a point and sigma is the surface charge density at that point. (The direction of hat n is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is (sigma )/(epsi_0) hatn. (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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Answer Text

Solution :
(a) Electric field on one side of a charged body is `E_(1)` and electric field on the other side of the same nody is `E_(2)`. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by, <br> `vec(E)_(1) = -(sigma)/(2 in_(0)) hat(n)` ....(i) <br> Where `hat(n)` = Unit vactor normal to the surface at a point <br> `sigma` = Surface charge density at that point <br> Electric field due to the other surface of the charged body, <br> `vec(E)_(2) = -(sigma)/(2 in_(0))hat(n)` ...(ii) <br> Electric field at anny due to the two surfaces, <br> `vec(E)_(2)-vec(E)_(1) = (sigma)/(2 in_(0)) hat(n)+(sigma)/(2 in_(0)) hat(n) = (sigma)/(in_(0))hat(n)` <br> `(vec(E)_(2)-vec(E)_(1)).hat(n) = (sigma)/(in_(0))` ...(iii) <br> Electric field at any point due to the two surfaces, <br> Since inside a closed conductor, `vec(E)_(1) = 0`, <br> `vec(E) = vec(E)_(2) = -(sigma)/(2 in_(0)) hat(n)` <br> Therefore, the electric field just outside the conductor is `(sigma)/(in_(0))hat(n)`. <br> (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.

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