What should be the electric field intensity inside a sphere

In summary, the electric field intensity and charge density inside a charged solid metallic sphere would remain the same as if the force acting between two point charges were proportional to \frac{1}{r^ 3}.
  • #1
PKM
49
16
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?
 
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  • #2
Good question. Makes one think. What do you think ?

PS if this is homework, then there is a special forum to post in.
 
  • #3
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.
 
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  • #4
##\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.
 
  • #5
Khashishi said:
##\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.
Khashishi said:
##\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 to What should be the electric field intensity inside a sphere

1. What is the formula for calculating the electric field intensity inside a sphere?

The formula for calculating the electric field intensity inside a sphere is E = Q/(4πεr^3), where E is the electric field intensity, Q is the charge enclosed by the sphere, ε is the permittivity of the material, and r is the radius of the sphere.

2. Does the electric field intensity inside a sphere depend on the material of the sphere?

Yes, the electric field intensity inside a sphere depends on the permittivity of the material. Materials with higher permittivity will have a higher electric field intensity inside the sphere.

3. How does the electric field intensity inside a sphere change with distance from the center?

The electric field intensity inside a sphere is directly proportional to the inverse of the distance from the center. This means that as the distance from the center increases, the electric field intensity decreases.

4. Can the electric field intensity inside a sphere be negative?

Yes, the electric field intensity inside a sphere can be negative. This occurs when the charge enclosed by the sphere is negative, resulting in an inward electric field towards the center of the sphere.

5. How does the size of the sphere affect the electric field intensity inside?

The size of the sphere does not affect the electric field intensity inside. As long as the charge enclosed by the sphere remains the same, the electric field intensity will also remain the same regardless of the size of the sphere.

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