
#1
Jan206, 03:18 PM

P: 129

What is the electric field inside a sphere of radius R when the charge density is given by:
[tex]\rho=\frac{A}{r}[/tex] Where A is a constant, and r is the radius at where the charge density is to be evaluated. By Gauss' law I have calculated the field to be equal to: [tex]E=\frac{A}{2\epsilon_0}[/tex] But that seems quite odd as the field is then independent of the radius. But could this be explained by the fact that the charge density is inversely proportional to the radius? Thanks! 



#2
Jan206, 03:41 PM

P: 308

That's correct. If you integrate the charge density over a sphere of radius R it's a little easier to see why this is so:
[tex]\int_0^R \int_0^{2 \pi} \int_0^{\pi} \rho r^2 \sin^2\theta d\theta d\phi dr[/tex] [tex]\int_0^R \int_0^{2 \pi} \int_0^{\pi} A r \sin^2\theta d\theta d\phi dr[/tex] [tex]A 4 \pi \int_0^R r dr[/tex] [tex]A 2 \pi R^2[/tex] So, the ammount of charge which is inside a sphere of radius R increases with the square of the radius. The surface area of that sphere which the electric field lines are spread out over also increases with R^2, although it's larger by a factor of two. This is why there is no R dependence on the electric field, the two competing effects of total charge and surface area over which the electric field is spread cancel out. 



#3
Jan306, 12:28 AM

P: 60

I'm just starting my second Calculus course, those nested integrals look fun




#4
Jan306, 01:42 AM

P: 308

Electric field inside a sphere of charge density A/r[tex]\int_0^{2 \pi} \int_0^{\pi} \sin^2\theta d\theta d\phi = 4 \pi[/tex] comes up all the time (such as in this problem). 



#5
May310, 10:04 PM

P: 1

This is good, except I'm pretty sure the volume element on that integral is r^{2} sin[theta] dr d[theta] d[phi] for spherical coordinates instead of having a sin^{2} [theta] component. This is from Griffith's Intro to E&M



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