About electrics
|dw:1474843237213:dw|
@ljetibo @Issy14 @SapphireMoon @osprey
So that means that the charge is actually evenly distributed
Because the electrons do not move correct.
i think that's right
so Gaussian Surface or what?
2R is outside the sphere
thnx for transcribing the question. Much easier this way. Yeah, someone would have to pick a shell as a gaussian surface and then do the integrals. I know that there will be 2 solutions right now: 1) when your r is bigger than R, then the electric field will be as if you're dealing with a point charge in the center of the sphere and 2) when your r is less than R, then electric field's gonna be something funky looking.
i'm seeing: linear for \(r <R\), inverse square for \(x>R\) with hindsight, not sure it helps on the original qu, though
yep, that was just BS !!
thank you very much
Join our real-time social learning platform and learn together with your friends!