Electrostatic Potential over all space

[Demon117]

If I have a sphere with radius R which has a charge distribution given by

$\rho(r)=\frac{5Q}{\pi R^{5}}r(r-R)$

and $\rho = 0$ at r bigger or equal to R, how do I find the electrostatic potential of this overall space? There is a charge Q, in addition, at the origin.

My original thought was to just do the usual and use

$V(r)=\frac{1}{4\pi\epsilon_{0}}\int \frac{\rho(r')}{r}dt'$,

which if I am correct the integration goes from 0 to R, correct. Or does it extend from infinity into R? This has never made much sense to me. Somebody help me out with this idea. Thanks!

 

[Demon117]

Or do I integrate from 0 to R, plus integrate from R to r? That seems a lot more in line with "over all-space". . . let me know your thoughts.

 

[mfb]

For the potential at a given radius r, you can neglect all charges outside (r'>r) and assume that all charges inside are at r'=0. This is similar to gravity, and follows from the spherical symmetry.

Therefore, for radius r, $\frac{dV(r)}{dr}=\frac{Q(r)}{r^2}$ with prefactors depending on your units. Q(r) is the total charge up to radius r: $Q(r)=Q_0 + \int_0^r 4 \pi r'^2 \rho(r') dr'$.
You can find an analytic expression for Q(r), this can be used in the first equation, and another integration will give you the potential.

 

[vanhees71]

I'd rather solve the differential equation (written in Heaviside-Lorentz units)
$$\Delta \Phi=-\rho.$$
Since the charge distribution is radially symmetric, you can make the ansatz in spherical coordinates,
$$\Phi(\vec{x})=\Phi(r).$$
Then from the Laplacian in spherical coordinates you get
$$\Delta \Phi=\frac{1}{r^2} [r^2 \Phi'(r)]',$$
and the equation becomes an ordinary differential equation, which you have to solve with the appropriate boundary conditions. This leads to mfb's solution (modulo a sign and prefactors depending on the system of units used).

 

[sardar]

I would approach this problem by first defining the problem and understanding the given information. We are dealing with a sphere of radius R that has a charge distribution described by \rho(r)=\frac{5Q}{\pi R^{5}}r(r-R), with an additional charge Q at the origin. We are looking to find the electrostatic potential over all space.

To find the electrostatic potential, we can use the formula V(r)=\frac{1}{4\pi\epsilon_{0}}\int \frac{\rho(r')}{r}dt', where r is the distance from the origin to the point where we want to calculate the potential. The integration goes from 0 to R because we are only concerned with the potential inside the sphere. This is because the charge distribution is defined as \rho = 0 at r bigger or equal to R, meaning there is no charge outside the sphere.

To better understand this concept, we can visualize the charge distribution as a solid sphere with a varying charge density. We can think of the charges inside the sphere as creating a potential that decreases as we move away from the center, and this potential becomes zero at the surface of the sphere.

To calculate the potential at a specific point, we need to take into account the contribution from all the charges inside the sphere. This is where the integral comes in - it allows us to sum up the contributions from all the charges, taking into account their distance from the point of interest.

In summary, to find the electrostatic potential over all space for this specific scenario, we would use the formula V(r)=\frac{1}{4\pi\epsilon_{0}}\int \frac{\rho(r')}{r}dt' with the integration going from 0 to R, since the charge distribution is only defined inside the sphere. I hope this helps clarify the concept for you.