Spherical Ball with uniform Charge Density

In summary: If you use the Gauss Law equation, you can find the electric field for any charge distribution, not just a point charge or sphere. In summary, we can use Gauss Law to find the electric field for any charge distribution, not just a point charge or sphere, by using the equation ##E = \frac{1}{4\pi\epsilon_0}\int_V\rho dV##, which takes into account the symmetry and uniform charge density of the system.
  • #1
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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
 
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  • #2
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##
 
  • #3
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
 
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  • #4
$$ E = \frac{Q}{4\pi r^2 \epsilon _0}$$

That's pretty general. It applies because that's the electric field for a sphere or point charge.
 
  • #5


Your equations for ##\Phi## and ##\vec{E}## seem to be correct for the general case of a spherical ball with uniform charge density. However, it is always a good practice to double check your work and make sure all the units and symbols are correct. Additionally, it may be helpful to provide some context or explanation for the equations, as well as any assumptions or simplifications that were made in the derivation. This can help others understand and verify your work more easily. Overall, your response appears to be well thought out and shows a good understanding of the topic.
 

1. What is a spherical ball with uniform charge density?

A spherical ball with uniform charge density is a theoretical model used in physics to describe a uniformly charged sphere. It is assumed that the charge is evenly distributed throughout the entire volume of the sphere.

2. How is the electric field inside and outside a spherical ball with uniform charge density calculated?

The electric field inside and outside a spherical ball with uniform charge density can be calculated using the Gauss's Law, which states that the electric field is directly proportional to the charge enclosed by a surface. In this case, the enclosed charge is the total charge of the sphere.

3. Does the electric field inside a spherical ball with uniform charge density depend on the distance from the center?

No, the electric field inside a spherical ball with uniform charge density is independent of the distance from the center. This is because the charge is evenly distributed throughout the sphere, resulting in a uniform electric field.

4. How does the electric potential change inside and outside a spherical ball with uniform charge density?

The electric potential inside and outside a spherical ball with uniform charge density can be calculated using the formula V = kQ/r, where k is the Coulomb's constant, Q is the total charge of the sphere, and r is the distance from the center. Inside the sphere, the potential decreases as the distance from the center increases, while outside the sphere, the potential decreases as the distance from the sphere's surface increases.

5. Can a spherical ball with uniform charge density exist in real life?

No, a spherical ball with uniform charge density is a theoretical model used for simplifying calculations in physics. In reality, it is difficult to achieve a perfectly uniform distribution of charge throughout a sphere. However, this model is still useful for understanding and predicting the behavior of charged spheres in various situations.

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