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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

`(vec"E"_2 - vec"E"_1).hat"n" = sigma/in_0`

Where `hat"n"` is a unit vector normal to the surface at a point and σ 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 σ `hat"n"/in_0`

**(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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#### Solution

**(a) **Electric field on one side of a charged body is E_{1} and electric field on the other side of the same body 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,

`vec"E"_1 = -sigma/(2in_0)hat"n"` ..........(1)

Where,

`hat"n"` = 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 charged body,

`vec"E"_2 = sigma/(2in_0)hat"n"` ........(2)

Electric field at any point due to the two surfaces,

`vec"E"_2 - vec"E"_1 = sigma/(2in_0)hat"n" + sigma/(2in_0)hat"n" = sigma/(in_0)hat"n"`

`(vec"E"_2 - vec"E"_1)hat"n" = sigma/in_0` ......(3)

Since inside a closed conductor, `vec"E"_1` = 0,

∴ `vec"E" = vec"E"_2 = -sigma/(2in_0)hat"n"`

Therefore, the electric field just outside the conductor is `sigma/(in_0)hat"n"`.

**(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 the electrostatic field is continuous from one side of a charged surface to the other.

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