Question

A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly 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 μ C/m2. What is the net charge in the shell?

Answer 1

dQ/dV = ρ

In other words a infinitesimal segment dV caries the charge
dQ = ρ dV

Let dV be a spherical shell between between r and (r + dr):
dV = (4π/3)·( (r + dr)² - r³ )
= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ )
= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ )
drop higher order terms
= 4·π·r²·dr

To get total charge integrate over the whole volume of your object, i.e.
from ri to ra:
Q = ∫ dQ = ∫ ρ dV
= ∫ri→ra { (b/r)·4·π·r² } dr
= ∫ri→ra { 4·π·b·r } dr
= 2·π·b·( ra² - ri² )

With given parameters:
Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² )
= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² )
= 3.77×10⁻⁸C
= 37.7nC

source: http://answers.yahoo.com/question/index?qid=20090702152524AA3IAAr

Answer 2

okay i want to show how i did it and want to know why im not getting the correct answer-

Answer 3

\[\rho = q/V\]

Answer 4

but \[\rho= b/r = 3/2/100 = 300/2\]
300/2 = q/V;
V = 4/3 pie [6^3-4^3]=608 *pie/3;
300/2= q/608*pie/3
therefore q = 95456 C..@stormfire1 and @Rohangrr help

Answer 5

@stormfire1

Answer 6

@goutham1995

I'll try to explain why you end up with a contradiction. The main reason is this:
Your notations are inadequate. The problem is you are calling with the same symbol \(\rho\) different things.

You should distinguish:
\( \rho\,(r)=\Large \frac{dQ}{dV}=\frac{b}{r}\) which is a function of r, representing how charge-density varies with distance from origin.

from:

\( \bar \rho=\Large \frac{Q}{V}\) which is the mean value of \(\rho \;(r)\) on your domain.

This \( \bar \rho\) can only be worked out when you have calculated Q in the first place.

In the light of what I've explained so far, you should understand why your:
\(\rho= b/r = 3/2/100 = 300/2\)
is 'nonsensical' ;-)

What you wrote next, using potential V, has no relation to the question asked.

I have not fully checked Rohangrr's answer, but this seems the right method.
personally, I do not go into so complicated considerations calculating dV.

I write dV = A(r) dr straight away, with A(r) = 4πr², area of the sphere on which the infinitesimal shell is built.

Answer 7

hey that V which i used was volume not potential

Answer 8

so \[\rho\] is not equal to 300/2?

Answer 9

Sorry about my mistake with V for volume! I was mislead by π as in \(\LARGE \frac {1}{4\pi \epsilon _0}\)

Which \(\rho\) are you talking about?
Where does 300 come from? Where does 2 come from?

\(\rho(r)\) varies continuously from \(\rho (R_1)=\Large \frac {b}{R_1}\) to \(\rho (R_2)=\Large \frac {b}{R_2}\)