anonymous
  • anonymous
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?
OCW Scholar - Physics II: Electricity and Magnetism
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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 \[\sigma/\epsilon _{0}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.