anonymous 5 years ago Another more specific question: In Module 3 in the course notes (p.2-25) is written that “Notice the discontinuity in electric field as we cross the plane. The discontinuity is given ...then the normal component of the electric field across that surface always exhibits a discontinuity with ΔEn=σ/ε0”. Could someone explain this please?

The charge lies in a plane where z = 0 (the x-y plane) Above the plane, in the positive z region, the electric field points away from positive charge, and so will be in the positive-z direction (if the sheet of charge is positive). On the other side of the plane of charge, in the negative z region, the electric field also points away from positive charge, and so must point in the negative z direction (if the sheet of charge is positive). That means that the electric field must go from pointing in the positive z direction just above the sheet of charge to pointing in the negative z direction just below the sheet of charge - that's a discontinuity. The notes show how the magnitude of the field was calculated ($\sigma/2\epsilon _{0}$) just above and below, but opposite in sign (direction) so that the total change in E is $\sigma/\epsilon _{0}$