A nonconducting spherical shell, with an inner radius of 4.0 cm and an outer radius of 6.0 cm, has c

A nonconducting spherical shell, with an inner radius of 4.0 cm and an outer radius of 6.0 cm, has charge spread nonuniformly through its volume between its inner and outer surfaces. The volume charge density ρ is the charge per unit volume, with the unit coulomb per cubic meter. For this shell ρ = b/r, where r is the distance in meters from the center of the shell and b = 3.0 µC/m^2. What is the net charge in the shell?

[mechengics]

Made our learning of Resnik Holiday Easier
Thank you very much sir

[paulbu]

actually we don't need to integrate. we know volume of sphere is (4/3 pi r^3) and the shell is just outer - inter radius. Then, we just times the density and get
charge q = b/r * (4/3 pi r^3) = b * 4/3 * pi * r^2 = 4/3 * b * pi* (a^2 - b^2)

[britbrits]

Thank you so much for this explanation! It was so helpful.

[larrysingian]

Very helpful! Appreciate you doing this for us :D

[maxwellacioli]

What is the eletric field in the shell?

[RayMysteryo]

so what if the charge was spread uniformly throughout the entire volume of the shell? does that mean that limits of integration would just be the entire radius of the shell?

[Nicolebedine]

i haven't even started watching but i love it already, God bless you for helping us, Thank youuu <3

[sujal...............]

Thanks 😊 it help ma a lot i dont understand how to find formula of this no i got it
This is a small part of jee advanced Q ☺️

[fettishferrubbish]

Hi. I have an integration question. In my college career, we are learning integration concurrently with calc-based physics. That is, I'm taking calc two while learning all about kinematics. Anyways, it seems like right now i'm seeing really simple integrals..where we can take out constants and just anti-derive a simple polynomial. What physics scenarios can you think of where we will be using trickier integration...like utilizing integration by parts? ...or even crazier..double...triple integrals? Thanks :)