Electrostatic field discontinuity illustrated

Question

I am trying to visually understand the discontinuity of the electric field when we move along a charged surface. If I am to draw the field lines below the surface and above, since we have a discontinuity of the normal component of the field and a continuity of the parallel component, then does this mean that the lines should have same direction/ angle with the surface (to keep the parallel component the same) while the length of the lines should be shorter in one side?

Answer 1

There are no field lines (electrostatics) inside an ideal conductor for the electric field is zero everywhere within, the whole thing is on the same potential. Just outside right on the conductor the tangential component of the field is zero, so it is indeed continuous across the boundary. If you integrate the field around a closed contour you get zero for $\nabla \times \mathbf E = 0$, that is $\oint_{\mathcal C} \mathbf E \cdot d\ell=0$ for any closed contour.

If you take two points on the boundary, say $\mathcal {P_1,P_2}$ and the contour has two segments $\mathcal {C_1,C_2}$, one outside and the other inside between these points then you will get $\int_{\mathcal {P_1}}^{\mathcal P_2}\mathbf E \cdot d\ell=0$ whether you integrate inside or outside; this just means that the potential between the points $\mathcal {P_1,P_2}$, both on the boundary, is zero.