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Introductory Physics Homework Help
Gauss' law, spherical symmetry
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[QUOTE="BvU, post: 5128721, member: 499340"] Time to review the relevant equations, all variables and given/known data: No idea what some of your variables stand for (you should list them in the problem statement), but if I guess: F=qE=ma ##\qquad## F = qE is interesting. Apparently we need E E=σ/2ξo ##\qquad##interesting too, but no business here. ##\sigma## is usually a surface charge density. No idea what ##\xi_0## stands for :rolleyes: a=σe/2εm ##\qquad##idem dito. ##\epsilon## times ##m## or ##\epsilon_m## ? Who wants a anyway ? [INDENT](don't take this as offensive: I just want to show that one has good reasons to be as clear as possible)[/INDENT] Leaves us with some curiosity wrt the "electric field: sphere of uniform charge" which of course you googled at the outset ? (And up pops hyperphysics, and the word Gauss shows up too !) We need ##\left | \vec E(r) \right | ## for ## r = R##, for some ##r_1 < R## and for some ##r_2 > R## For ## r = R## you made a start. You get $$F_R={\sigma q \over 4π\epsilon_0 R^2}\ ,$$which does not have the dimension of a force. Can't be right. [INDENT]what is ##\sigma## ? ##q## ? dimensions ? [/INDENT] Nor is $$F_R={\sum q \over 4π\epsilon_0 R^2}\ ,$$ which does not have the dimension of a force either. But at least this has the dimension of an electric field strength ! The very ##\left | \vec E(R) \right | ## we were after ! So what is ##\left | \vec E(r) \right | ## for ## r ## if ##r < R## ? And what is the expression for ##r > R## ? [/QUOTE]
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Gauss' law, spherical symmetry
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