Gauss Law and a hollow spherical shell

[stunner5000pt]

A hollow spherical shell carries a charge density
$$\rho = \frac{k}{r^2}$$

in the region a<= r <= b. As in the figure
Find the elctric field in these three regions
i) r <a
ii) a<r<b
iii) r>b

SOlution:
for r<a it simple.. no exclosed charge for any gaussian sphere within that region so E = 0

for a<r<b
the thing which stumps is the charge density... isn't charge density given in coulombs per cubic metre usually??

what I am concerned about in the enclosed charge in teh gaussian sphere of radius a<r<b... would i integrate rho from r' = a to r' = r??

rho dot dr would give the total charge enclosed, no??

doing that gives
$$q_{enc} = k\left(\frac{1}{a} - \frac{1}{r}\right)$$

[tex] E (4 \pi r^2) = \frac{k}{epsilon_{0}} \left(\frac{1}{a} - \frac{1}{r}\right)

for the third part that is r>b would i do something similar but integrate from r' =a to r'=b??

Please help!

Help is always greatly appreciated!

 

[Doc Al]

for a<r<b
the thing which stumps is the charge density... isn't charge density given in coulombs per cubic metre usually??

Absolutely. Which means that k has what units?

what I am concerned about in the enclosed charge in teh gaussian sphere of radius a<r<b... would i integrate rho from r' = a to r' = r??

Yes.

rho dot dr would give the total charge enclosed, no??

No! An element of charge would be rho dV, not rho dr. rho is the charge density per unit volume.

 

[stunner5000pt]

Absolutely. Which means that k has what units?

No! An element of charge would be rho dV, not rho dr. rho is the charge density per unit volume.

so K has units of C/m

so i should be integrating rho dV but still r'=a to r'=r

thanks for the help

 

[Doc Al]

so K has units of C/m

so i should be integrating rho dV but still r'=a to r'=r

Yes and yes.