What should be the electric field intensity inside a sphere

PKM
49
16

Main Question or Discussion Point

If the force acting between two point charges were proportional to [itex]\frac{1}{r^ 3}[/itex], instead of [itex]\frac{1}{r^ 2}[/itex], what would be the electric field intensity and charge density inside a charged solid metallic sphere?
 

Answers and Replies

BvU
Science Advisor
Homework Helper
12,557
2,832
Good question. Makes one think. What do you think ?

PS if this is homework, then there is a special forum to post in.
 
Khashishi
Science Advisor
2,815
492
Hmm. The charge distribution will settle into a minimum energy configuration. For ##\frac{1}{r^2}##, all of the charge would go to the surface of the sphere, and the field inside would be zero. For ##\frac{1}{r^3}##, the field inside would still have to be zero (otherwise, charges would continue to redistribute) but it isn't so obvious what the charge density is.
We need to find a distribution such that the field is 0. By symmetry, the distribution must be only a function of radius.
Pick an arbitrary point in the sphere. Without loss of generality, we assume the point to be on the z-axis, going through the center of the sphere. Let the position be ##(0,0,z')##. The field at that point is
##0 = F = \int\limits_0^\pi \int\limits_0^R 2\pi f(r,\theta) r^2 \sin \theta dr d\theta##
where ##f(r,\theta)## is the piece of field at the point due to the charge density at ##(r,\theta)##. We can ignore the ##\phi## coordinate due to symmetry.
The amplitude of the piece of field f is
##|f(r,\theta)| = \rho(x,y,z) \frac{1}{[(x-x')^2 + (y-y')^2 + (z-z')^2]^{3/2}}##
where (x,y,z) are the position of the differential charge and (x',y',z') = (0,0,z') is the arbitrary point.
##z = r \cos \theta##
##x^2 + y^2 = (r \sin \theta)^2##
so
##|f(r,\theta)| = \rho(x,y,z) \frac{1}{[(r \sin \theta)^2 + (r \cos \theta -z')^2]^{3/2}}##
The angle ##\alpha## of the field f to the z axis is given by
##\tan \alpha = \frac{r \sin \theta}{r \cos \theta - z'}##
so the z component of the field f is
##f_z = \cos \alpha |f(r,\theta)|##
to be continued.
 
Last edited:
Khashishi
Science Advisor
2,815
492
##\cos \alpha = \sqrt{\frac{(r \cos \theta - z')^2}{(r \sin \theta)^2 + (r \cos \theta - z')^2}}##
##f_z(r,\theta) = \rho(r) \frac{r \cos \theta - z'}{[(r \sin \theta)^2 + (r \cos \theta - z')^2]^2}##
Now go back to the first equation, setting ##F_z=0##
##0 = F_z = \int\limits_0^\pi \int\limits_0^R 2\pi f_z(r,\theta) r^2 \sin \theta dr d\theta##
Now, it should be possible to integrate ##f_z## over ##\theta##. This gives an integral equation for ##\rho(r)##
I guess you can then take the derivative to get an equation for ##\rho(r)##.
This approach seems messy, and I wonder if there's a better way.
 
PKM
49
16
##\cos \alpha = \sqrt{\frac{(r \cos \theta - z')^2}{(r \sin \theta)^2 + (r \cos \theta - z')^2}}##
##f_z(r,\theta) = \rho(r) \frac{r \cos \theta - z'}{[(r \sin \theta)^2 + (r \cos \theta - z')^2]^2}##
Now go back to the first equation, setting ##F_z=0##
##0 = F_z = \int\limits_0^\pi \int\limits_0^R 2\pi f_z(r,\theta) r^2 \sin \theta dr d\theta##
Now, it should be possible to integrate ##f_z## over ##\theta##. This gives an integral equation for ##\rho(r)##
I guess you can then take the derivative to get an equation for ##\rho(r)##.
This approach seems messy, and I wonder if there's a better way.
##\cos \alpha = \sqrt{\frac{(r \cos \theta - z')^2}{(r \sin \theta)^2 + (r \cos \theta - z')^2}}##
##f_z(r,\theta) = \rho(r) \frac{r \cos \theta - z'}{[(r \sin \theta)^2 + (r \cos \theta - z')^2]^2}##
Now go back to the first equation, setting ##F_z=0##
##0 = F_z = \int\limits_0^\pi \int\limits_0^R 2\pi f_z(r,\theta) r^2 \sin \theta dr d\theta##
Now, it should be possible to integrate ##f_z## over ##\theta##. This gives an integral equation for ##\rho(r)##
I guess you can then take the derivative to get an equation for ##\rho(r)##.
This approach seems messy, and I wonder if there's a better way.
I appreciate your method, thanks. The point I wanted to clarify is, by intution, to understand whether the charge density inside the sphere would drop to zero. Clearly, it wouldn't.
You have proceeded to a clear mathematical treatment, which is, very excellent of you.
 

Related Threads for: What should be the electric field intensity inside a sphere

  • Last Post
Replies
9
Views
1K
Replies
1
Views
9K
Replies
16
Views
2K
  • Last Post
Replies
1
Views
543
Replies
8
Views
1K
Replies
5
Views
2K
Replies
32
Views
3K
Replies
2
Views
1K
Top