0
$\begingroup$

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 $\sigma = \sigma_0 cos(\theta)$

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $\rho dV = \sigma_0 \cos(\theta)dS$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.