Question

two 8cm diamter charged disks face each other, 25cm apart. both disks are charged to -60nC. what is the electric field strength a point on the axis 5cm from one disk between them?

Answer 1

The field strength between the two discs is uniform and given by \( \sigma / \epsilon_0 \) where \( \sigma \) is the charge density on the plates.

Answer 2

what i have is...
E=kq/r^2
and i plug numbers in, came out to be negative cuz i have the direction as going left and then added E_right...is this right?

Answer 3

did i use the wrong formula? i've been trying to find the formula for field between charged disks...

Answer 4

what i used didn't involve charge density

Answer 5

Yes, you need the formula for a equipotential plane. Let me find this for you ....

Answer 6

ok thx

Answer 7

Look at this lecture, beginning at 23:30
http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-3-electric-flux-and-gausss-law/

Answer 8

or 23:50

Answer 9

ok and do you mean when considering uniform elec field, charge density should be involved in the equation?

Answer 10

he develops the formula I cited above. If that formula isn't familiar, you should watch this so you begin to develop the right kind of intuition here.

Answer 11

very nice lecture, unfortunately i didn't catch anything about disks. it's definitely not like an infinitely large plane. he may explain it later in the video but i will have to watch another time. too much to study, so little time.

Answer 12

and what A difference in the quality of education......MIT

Answer 13

Actually, as the plates have the same charge, to first approximation, the charge between the plates is zero. Outside the plates the charge is \( \sigma/\epsilon_0 \).

I take your point about the plates not being infinite, but I suggest to you that this is what your question wants to use as an approximation. Whatever else, the Coulomb field formula you've written down is not a good approximation.

Answer 14

charge or field? outside the plate, the field is charge density/epsilon_naught. and if you are talking about field, isn't it only at the center where it equals zero? since left and right E_field cancels each other?

Answer 15

Yes, technically, only in the center. But using the approximation of the plates being large, then the field is everywhere uniform.