Spherical Ball with uniform Charge Density

AI Thread Summary
The discussion focuses on calculating the electric potential (Φ) and electric field (E) for a uniformly charged spherical ball using complex integrals. One participant suggests simplifying the approach by applying Gauss's Law due to the symmetry of the problem, arguing that the convoluted equations are unnecessary. They clarify the distinction between electric potential and electric flux, emphasizing that Gauss's Law can yield the results more straightforwardly. The general case is addressed by referencing the electric field formula for a sphere or point charge, which simplifies the calculations significantly. Overall, the conversation highlights the importance of leveraging symmetry in electrostatics to simplify complex problems.
Philosophaie
Messages
456
Reaction score
0
I want to find ##\Phi## and ##\vec{E}## for the general case of a Spherical Ball with uniform Charge Density centered at the origin radius d.

##\Phi = \frac{\rho}{4*\pi*\epsilon_0}\int\int\int\frac{r^2*sin\theta}{|r-r'|}dr d\theta d\phi##
##E = \frac{\rho}{4*\pi*\epsilon_0}\int\int\int\frac{r^2*sin\theta}{|r-r'|^2}*e_{r-r'}*dr d\theta d\phi##

For the General Case of Point {a, b, c}:

|r-r' | ^ 2 = (x-a) ^2 + (y-b) ^2 + (z-c) ^2
= r ^2 + a ^2 + b ^2 + c^2 -2r*##(a*cos\theta*cos\phi+b*sin\theta*cos\phi+c*sin\phi)##

Could someone check my work?

##\Phi = \frac{\rho}{4*\pi*\epsilon_0}## ## \int\int\int \frac{r^2*sin\theta}{\sqrt{r^2 + a^2 + b^2 + c^2 -2r(a*cos\theta*cos\phi+b*sin\theta*cos\phi+c*sin\phi)}}dr d\theta d\phi##

##E = \frac{\rho}{4*\pi*\epsilon_0}## ##\int\int\int\frac{r^2*sin\theta*\vec{r}}{( r^2 + a^2 + b^2 + c^2 -2r(a*cos\theta*cos\phi+b*sin\theta*cos\phi+c*sin\phi))^{3/2}} dr d\theta d\phi##

Those who wish to appear wise among fools, among the wise seem foolish.
- Quintilian
 
Last edited:
Physics news on Phys.org
I'm not really sure what you're doing, your work is very convoluted.

just use Gauss's Law.

$$\oint _S \mathbf E \cdot d\mathbf S = \frac{Q}{\epsilon _0}$$

You don't need to use the more complicated equations to solve this because we have symmetry and the charge density is uniform.

And it's not clear by what you mean with ##\Phi##. I've seen that referring to either the electric potential and also the electric flux.

For electric flux, $$\int _S \mathbf E \cdot d \mathbf S = \Phi$$ and if you want the electric potnetial, it can be found by ##\mathbf E = - \nabla \Phi##
 
Gauss Law because of symmetry becomes:

##E= \frac{\int_ V \rho dV}{\epsilon_0\int_S dS}##

##E = \frac{\rho*4/3*\pi*d^3}{4*\pi*r^2}##

How does this apply to the general case above?


Wisdom is not a product of schooling but of the lifelong attempt to acquire it.
- Albert Einstein
 
Last edited:
$$ E = \frac{Q}{4\pi r^2 \epsilon _0}$$

That's pretty general. It applies because that's the electric field for a sphere or point charge.
 
Thread 'Motional EMF in Faraday disc, co-rotating magnet axial mean flux'
So here is the motional EMF formula. Now I understand the standard Faraday paradox that an axis symmetric field source (like a speaker motor ring magnet) has a magnetic field that is frame invariant under rotation around axis of symmetry. The field is static whether you rotate the magnet or not. So far so good. What puzzles me is this , there is a term average magnetic flux or "azimuthal mean" , this term describes the average magnetic field through the area swept by the rotating Faraday...
Back
Top