Non-conducting sphere as a point charge kept at its centre (Is it valid assumption?)

Question

Suppose we have two isolated non-conducting spheres with charges say, $Q_1$ and $Q_2$ respectively, and their centres are separated by a distance of $R$, to calculate the forces between them, we just do

$F=\frac{k.Q_1.Q_2}{R^2}.$

Basically we assume them to be point charges kept at their centre and then calculate the forces of those point charges. Now, I know electric field of any of the spheres resembles that of a point charge but for finding the force it's not obvious to assume the other sphere to be a point charge, and multiply the electric field with the other sphere's charge. Similar thing we do while calculating the potential energy of the system.

How can we assume the spheres to be point charges for any practical purposes? Is it just the result that we get after NOT assuming them to be point charges and calculating the force through integration?

Answer 1

When we know that the electric field outside a uniformly charged sphere is the field of a point charge placed in the center of the sphere, it is sufficient to use Newton's Third Law to prove that the two spheres act on each other with the same forces as two point charges. There is no need to consider integrals, this is an exercise in logic. Indeed, the first sphere acts on the second sphere with the same force as a point charge in the center of the first sphere. Consequently, the second sphere acts on the first sphere with the same force with which it acts on a point charge in the center of the first sphere. But the second sphere acts on a point charge as a point charge in its center.

Update. After the questioner's comment I decided to give a more detailed answer. We will consider the second charged sphere as a set of infinitesimal charges $e_i$ with coordinates $\vec{r}_i$. Then the force with which the first sphere acts on the second sphere is equal to $$ \vec{F}_{12} = \sum_i e_i \vec{E}_i^{(1)}, $$ where $\vec{E}_i^{(1)}$ is the electric field created by the first sphere at the point $\vec{r}_i$. In the case when the center of the first sphere is in the coordinate center, we have $$ \vec{E}_i^{(1)} = \frac{kQ_1\vec{r}_i}{|\vec{r}_i|^3} $$ and accordingly $$ \vec{F}_{12} = Q_1\sum_i \frac{ke_i\vec{r}_i}{|\vec{r}_i|^3} = -Q_1 \vec{E}_0^{(2)}, $$ where $$ \vec{E}_0^{(2)} = -\sum_i \frac{ke_i\vec{r}_i}{|\vec{r}_i|^3} $$ is the electric field created by the second sphere in the center of the first sphere. In the case when the center of the second sphere is at the point $\vec{R}$, we have $$ \vec{E}_0^{(2)} = -\frac{kQ_2\vec{R}}{|\vec{R}|^3} $$ After all we obtain $$ \vec{F}_{12} = \frac{kQ_1Q_2\vec{R}}{|\vec{R}|^3} $$

Answer 2

It is a general property of electric fields that the average electric field over the surface of any sphere, due to charges outside that sphere, is equal to the value of the field at the center of the sphere. In other words, if we take the origin to be the center of the sphere and there are no charges inside the sphere of radius $R$, then $$ \frac{1}{4 \pi R^2} \iint \vec{E} \, \mathrm{d}a = \vec{E}(\vec{0}). $$ (The proof is a standard exercise in upper-level electrodynamics courses so I won't include it here.)

Given this fact, then the property you've described regarding the forces follows immediately, since the force on a uniform spherical shell of charge due to these external fields would be $$ \vec{F} = \iint \sigma \vec{E} \, \mathrm{d}a = 4 \pi R^2 \sigma \vec{E}(\vec{0}) = Q \vec{E}(\vec{0}) $$ where $Q = 4 \pi R^2 \sigma$ is the total charge of the shell.