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So I fundamentally lack understanding of electrostatics. Read!

B

beer

Lifer
For some reason, the explanation given to this problem just, does, not, make, sense!

Here is the problem:

Text

I understand everything up to:

"Gauss's Law tells us tha tsince we enclose no charge, there is no electric field due to the outer shell inside the inner shell." Which makes sense since all the charge is concentrated on the surface of the shell.

However, then it says:

"If there is no electric field, the potential stays the same as we go inwards. Therefore, the potential due to the outer shell at any point inside the outer shell is...."

My question is, why is there that contribution of q2/3a if we are inside the outer shell and there there is no electric field due to the outer shell?
 
The last statement before the equation for V2 should read, "Therefore the potential due to the inner shell at any point inside the outer shell is"

 
Originally posted by: KMurphy
The last statement before the equation for V2 should read, "Therefore the potential due to the inner shell at any point inside the outer shell is"

Just a simple frustrating typographical error.
Click to expand...

no, the statement is correct
 
Originally posted by: WinkOsmosis
What is your major? Homework server is leet.
Click to expand...

The HW server is the greatest abuse of professor power I have ever seen.

It allows them to assign tremendous amounts of homework without regard to how long it takes them to do. None of my EE classes use it because it just doesn't work for the material in question. If a physics prof had to have graders grade it, no way would they be assigning 90 minutes of homework, three times a week. Plus, since the the HW service is used for tests, it allows them to abuse that too. The homework service's deficiency is that it is all-or-nothing. Physics should not be all-or-nothing. If it takes me 6 steps to solve a problem, and I get steps 1-5 right, I get no points from the HW service. Pisses me off SO much.
 
i hate the hw server crap....CAPA its called here

ummmmm we never did anything with the inside of an outer shell we just dealth with outer and inner shells is that the same? (assuming the inner shell is the inner part of the outer shell)
 
im not fat im just big boned.....and since ive been sick since monday and couldn't eat much cuz my throat has swollen to half its original size i lost like 5 lbs......ye haw down to 132
 
the E field inside the outer shell is 0 no matter what is inside the shell, whether its totally solid, totally hollow, or somewhere in between. if it's totally hollow, the potential on the surface is k*q/3a and the potential inside is constant (i.e. the same as the potential on the surface at any point inside). so if you put something inside the hollow center, the potential is then the potential at that point from whatever you just put in the cavity PLUS the existing potential that would be there if the cavity was empty.
 
the explanation was only for the charge due to the outer shell. however, at point p, there is also a charge due to the inner shell.

charge due to outer shell: Vp = (1/(4*pi*e0)) * (q2/3a)

charge due to inner shell: Vp = (1/(4*pi*e0)) * (q1/2a)

since the charges would add at point P, you add them: Vp = (1/(4*pi*e0)) * ((q1/2a) + (q2/3a))

easy as that.


ebaycj
 
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