Ask your own question, for FREE!
Physics 54 Online
OpenStudy (blackstreet23):

A solid insulating sphere of radius R has a non-uniform charge density that varies with the distance r from the center according to the formula p(r) = p0(exp)(-r/b)/r^2 where b and p0 are constants. A)What type of Gaussian surface would you use in this problem? B) Derive an expression for the electric flux trough the Gausssian surface in terms of the electric field strength E(r). C) Derive an expression for E(r) a distance r < R from the sphere’s center in terms of the constants p0 and b. Note: The volume element for a sphere shell of radius r and thickness dr is dV = 4(pi)(r^2)dr.

OpenStudy (blackstreet23):

The gaussian surface that is convenient to use would be a sphere, which can be adapted to the conditions of the charge density given in the problem.

OpenStudy (blackstreet23):

OpenStudy (blackstreet23):

Wrong pics

OpenStudy (blackstreet23):

OpenStudy (blackstreet23):

OpenStudy (blackstreet23):

This is what I did with my classmates

OpenStudy (blackstreet23):

is that correct?

OpenStudy (blackstreet23):

@aliqanber

OpenStudy (blackstreet23):

@aaronq , @welshfella

OpenStudy (blackstreet23):

@Elsa213

OpenStudy (aliqanber):

I think your solution looks good.

OpenStudy (blackstreet23):

Should b be E(r)A = Qin/ epsalon0 ?

OpenStudy (blackstreet23):

like this but inversed? https://upload.wikimedia.org/math/6/a/8/6a81cca2936e2b38422f89853918e8de.png

OpenStudy (blackstreet23):

I am just nor sure because part b says derive and I am just using the formula that the book gives.

OpenStudy (aliqanber):

I think the answer for b is just flux = integral E dA = E (4*pi*r^2)

OpenStudy (blackstreet23):

and then you are 100% that the rest is fine?

OpenStudy (aliqanber):

yeah pretty sure.

OpenStudy (blackstreet23):

Thanks then @aliqanber appreciate your hard work :D

OpenStudy (aliqanber):

np

OpenStudy (irishboy123):

got same

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!