Question

a)Show that the normal component of electrostatic field has adiscontinuity from one side of a charged surface to anothergiven by (E2-E1).^n=σ/ϵwhere ˆn is a unit vector normal to the surface at a point ands is the surface charge density at that point. (The direction ofˆnis from side 1 to side 2.) Hence, show that just outside aconductor, the electric field is s ˆn /e0.b) Show that the tangential component of electrostatic field iscontinuous from one side of a charged surface to another. [Hint:For (a), use Gauss’s law. For, (b) use the fact that work done byelectrostatic field on a closed loop is zero.]

Answer 1

a)

The electric field due to one surface of the charged body is given as,

E → 1 =− σ 2 ε 0 n ^ (1)

Where, electric field on one side of a charged body is E → 1 , the unit vector normal to the surface at a point is n ^ and the surface charge density is σ.

The electric field due to other surface of the charged body is given as,

E → 2 =− σ 2 ε 0 n ^ (2)

Where, electric field on other side of a charged body is E → 2 .

By subtracting equation (1) from equation (2), we get

E → 2 − E → 1 = σ 2 ε 0 n ^ −( − σ 2 ε 0 n ^ ) = σ 2 ε 0 n ^ + σ 2 ε 0 n ^ = σ ε 0 n ^ (3)

Inside a closed conductor the electric field is zero so E → 1 =0.

By substituting the given values in the above expression, we get

E → 2 =− σ 2 ε 0 n ^

Thus, the electric field just outside the conductor is σ ε 0 n ^ .

b)

The work done by the electrostatic field will be zero if a charged particle is moved from one point to the other point on a closed loop. Due to this reason, the tangential component of electrostatic field is continuous from one side of a charged surface to the other side.