Question

Show that the tangential component of E
is continuous across a dielectric boundary.

Answer 1

Lets choose rectangular path with height h and lenght l , shown on picture.

From Faraday's law:

$$\oint \vec{E}\cdot \vec{dl} = 0$$

Assuming we made our rectangular small enough that E is roughly constant, and h significantly smaller than l,

$$\oint \vec{E}\cdot \vec{dl} = \int_1^2 \vec{E}\cdot \vec{dl}+ \int_2^3 \vec{E}\cdot \vec{dl}+ \int_3^4 \vec{E}\cdot \vec{dl}+ \int_4^1 \vec{E}\cdot \vec{dl} = E_{1t}l -E_{2t}l=0$$

Therefore,

$$E_{1t}=E_{2t}$$