I am reading the Feynman courses. It is written inside the book that a sphere of radius $a$ charged by a surface charge $\sigma = \sigma_0cos(\theta)$ should generate an electric field with a moment of $p = \frac{4\pi \sigma_0 a^3}{3}$ at the exterior of the sphere, and inside the sphere, the field should be constant and shall be $\vec{E}=-\frac{\sigma_0}{3\epsilon_0}\vec{z}$.

In the book, it is noted to use 2 spheres uniformly volumic charged (the spheres are charged in an opposite way). Inside the intersection of the two spheres, we got a zero charge: $\rho_{total} = \rho_1 + \rho_2 = \rho_+ + \rho_- = 0$ However, outside the intersection, I "guessed" that the electric charge is something like $\rho \cdot d\cdot cos(\theta)$ (When I say I guessed, I red it here : https://physics.stackexchange.com/a/489351/318644 ). With this result, I succeed to show the good result for the moment and the electric field.

However, I don't know how can I really demonstrate this result in a rigorous way... I am not even sure to understand it properly. So, if someone could explain me the "demonstration" that the two spheres uniformly charged in a volumic way are similar to a surfacic charged sphere of $\sigma = \sigma_0 cos(\theta)$


1 Answer 1


Okay, finally I think I got the idea. Here is the equation of a sphere of radius $R$:

$$R^2 = (x - a)^2 + (y - b)^2 + (z - c)^2$$ With $(a, b, c)$ the center of the sphere.

Let's say we take our 2 spheres and apply a displacement of $\frac{d}{2}$ on the z axis. We got for the positive and the negative sphere

$$a^2 = x^2 + y^2 + \left(z - \frac{d}{2}\right)^2\\a^2=x^2 + y^2 + \left(z + \frac{d}{2}\right)^2$$

Now, we want to compute the distance to the origin of both spheres' surfaces.

Let's take a rayon in any direction and use an intersection method to compute the distance to both spheres' surfaces.

$$\vec{r} = \left(\begin{array}{c}a + tx\\b+ty\\c+tz\end{array}\right)$$

Since we work at the origin, $(a, b, c) = \vec{0}$ and since we work with spherical coordinates, we got

$$\vec{r} = \left(\begin{array}{c} t\cdot \sin(\theta)\cdot \cos(\phi)\\ t\cdot \sin(\theta)\cdot \sin(\phi)\\ t\cdot \cos(\theta) \end{array}\right)$$

Let's replace these values into our prior equations.

$$a^2 = t^2 \cdot \cos^2(\phi) \cdot \sin^2(\theta) + t^2\cdot \sin^2(\phi)\cdot \sin^2(\theta) + t^2\cdot \cos^2(\theta) - d\cdot t\cdot \cos(\theta) + \frac{d^2}{4}$$

That gives $$t^2 - d\cdot t \cdot \cos(\theta) + \frac{d^2}{4} - a^2 = 0$$

We deduce that $\Delta = d^2\cdot \cos^2(\theta) - 4(\frac{d^2}{4} - a^2)$ Hence, the distance (positive) is $$t=\frac{d\cdot \cos(\theta) + \sqrt{d^2\cdot \cos^2(\theta) - 4(\frac{d^2}{4} - a^2)}}{2}$$

If we solve in the same manner for the second sphere, we got:

$$t=\frac{-d\cdot \cos(\theta) + \sqrt{d^2\cdot \cos^2(\theta) - 4(\frac{d^2}{4} - a^2)}}{2}$$

And so, the distance between the shell of the first sphere and the second sphere is $d\cdot \cos(\theta)$

So, we can deduce that (if $d$ ​is small enough):

$$\rho dV = \rho \cdot d\cdot \cos(\theta)dS$$ Taking $\rho = \sigma_0 / d$

We obtained as wanted $\rho dV = \sigma_0 \cos(\theta)dS$


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