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Total charge of a sphere

  • Thread starter Uku
  • Start date
  • #1
Uku
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Homework Statement


I am given a charge density for a solid sphere
[tex]\rho=14.1\frac{pC}{m^{3}}\frac{r}{R}[/tex]
The r is the distance from the center of the sphere and R is the radius of the whole thing.

[tex]R=5,6cm[/tex]

Now I am asked for the whole charge contained by the sphere.


Homework Equations



[tex]\rho=\frac{dq}{dV}[/tex]

The Attempt at a Solution


[tex]dq=\rho dV[/tex]
[tex]dq=4.1\frac{pC}{m^{3}}\frac{r}{R} dV[/tex]
I'll just denote the picocoulomb into B
[tex]q=\frac{B}{R} \int r dV[/tex]

Right, here I land. This is from Halliday, second year thing, I bet they don't expect you to do volume integration in spherical coordinates or anything such. I could write it:
[tex]dV=\frac{4}{3} \pi dr^{3}[/tex]?

Pff...

EDIT:

Ok, now I get it I think:
[tex]q=\frac{B}{R} \int r dV[/tex]
is actually
[tex]q=B \int dV[/tex]
[tex]q=14.1\frac{pC}{m^{3}} \frac{4}{3} \pi r^{3}[/tex]

ought to give me the right answer

EDIT:

It does not.
The right answer is given by
[tex]q=14.1\frac{pC}{m^{3}} \pi r^{3}[/tex]

But how do I land that?
 
Last edited:

Answers and Replies

  • #2
1,137
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just consider a small spherical shell of radius x(<R) and thickness dx

write dq for it (by finding volume of "shell" --- not of sphere of radius x----) and integrate it from x=0 to x=R
 

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