Spherical Ball with uniform Charge Density

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Discussion Overview

The discussion revolves around calculating the electric potential ##\Phi## and electric field ##\vec{E}## for a spherical ball with uniform charge density centered at the origin. Participants explore different approaches, including direct integration and the application of Gauss's Law, while addressing the implications of symmetry in the problem.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents integrals for calculating the electric potential and electric field, expressing concern about the complexity of their approach.
  • Another participant suggests using Gauss's Law due to the symmetry of the problem, arguing that it simplifies the calculations significantly.
  • A third participant reiterates Gauss's Law in a different form, questioning how it applies to the general case presented initially.
  • A fourth participant states a general formula for the electric field of a sphere or point charge, implying its relevance to the discussion.
  • There is ambiguity regarding the term ##\Phi##, as it may refer to either electric potential or electric flux, leading to confusion among participants.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate method for solving the problem, with some advocating for direct integration and others favoring Gauss's Law. The discussion remains unresolved regarding the best approach and the interpretation of ##\Phi##.

Contextual Notes

The discussion highlights the complexity of the problem and the potential for multiple interpretations of terms like ##\Phi##. There are unresolved aspects regarding the assumptions made in the integration approach and the application of Gauss's Law.

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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:
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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.
 

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