# What should be the electric field intensity inside a sphere

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## Main Question or Discussion Point

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

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Good question. Makes one think. What do you think ?

PS if this is homework, then there is a special forum to post in.

Khashishi
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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Khashishi
$\cos \alpha = \sqrt{\frac{(r \cos \theta - z')^2}{(r \sin \theta)^2 + (r \cos \theta - z')^2}}$
$f_z(r,\theta) = \rho(r) \frac{r \cos \theta - z'}{[(r \sin \theta)^2 + (r \cos \theta - z')^2]^2}$
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.
$\cos \alpha = \sqrt{\frac{(r \cos \theta - z')^2}{(r \sin \theta)^2 + (r \cos \theta - z')^2}}$
$f_z(r,\theta) = \rho(r) \frac{r \cos \theta - z'}{[(r \sin \theta)^2 + (r \cos \theta - z')^2]^2}$
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.