Question

i) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
$(E_{2} - E_{1}) . \hat{n} = \frac{σ}{ε_{0}}$
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 $σ \hat{n} / ε_{0} .$

ii) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: Use the fact that work done by electrostatic field on a closed loop is zero.]

Solution

Consider that E₁ and E₂ be the electric field on the two sides of the charged surface and XY be the charged surface of dielectric as shown in the figure below.
Note: Side 1 is the left side and side 2 is the right side.

i) Near a plane sheet of charge, electric field is given as-

E = \frac{σ}{2 ε_{0}}

Electric field on side 2 if

\hat{n}

is a unit vector normal to the sheet from side 1 to side 2-

E_{2} = \frac{σ}{2 ε_{0}}

(In the outward direction normal to the side 2)
Now, electric field on side 1 is-

E_{1} = \frac{σ}{2 ε_{0}}

(In the outward direction normal to the side 1)

As E₁ and E₂ are in opposite directions so will have opposite sign

∴

(E_{2} - E_{1}) \hat{n} = \frac{σ}{2 ε_{0}} - (- \frac{σ}{2 ε_{0}}) = \frac{σ}{ε_{0}} \hat{n}

There must be discontinuity at the sheet of charge since E₁ and E₂ act in opposite directions.
Now, electric field inside the conductor vanishes.
Hence, E₁ = 0
Therefore, electric field outside the conductor

E = E_{2} = \frac{σ}{ε_{0}} \hat{n}

ii) Consider a rectangular loop ABCD of length l and negligible small breadth b.
Line integral along the closed path ABCD will be

\oint E . dl = E_{1} . l - E_{2} . l = 0 \Rightarrow E_{1} . l \cos θ_{1} - E_{2} . l \cos θ_{2} = 0 (E_{1} \cos θ_{1} - E_{2} \cos θ_{2}) . l = 0 (E_{1}' - E_{2}') = 0

where, E₁' and E₂' are the tangential components of E₁ and E₂ respectively.
Hence, E’₁ = E’₂
Therefore, the tangential component of the electrostatic field is continuous across the surface.

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