Finding Electric Field using Gauss's Law for a Solid Ball with Charge Density

In summary, the electric field at a distance r < rb from the center of a sphere of charge density \rho is given by E*(4*pi*r_b^2) .
  • #1
Seraph404
64
0

Homework Statement



We have solid ball of radius rb and charge density [tex]\rho[/tex].

I need to find the magnitude of electric field E(r) at a distance r < rb from the center of the ball and express my answer in terms of [tex]\rho[/tex], r, rb, and [tex]\epsilon[/tex] 0.


Homework Equations



Gauss's Law
[tex]\int[/tex] EA = Q/[tex]\epsilon[/tex]0

The Attempt at a Solution




E*(4*[tex]\pi[/tex]*rb2) = 4*[tex]\pi[/tex]*r3*[tex]\rho[/tex]/(3*[tex]\epsilon[/tex]0)

E(r)= (r3*[tex]\rho[/tex])/(3*rb2*[tex]\epsilon[/tex]0)

But the website I check my answers with tells me it's wrong. Can someone help me to understand, please? The hint it gave me was that my answer should not depend on rb. Why wouldn't it?
 
Physics news on Phys.org
  • #2
This looks right on first glance.
The integral on the left should yield

E*(area enclosed by a sphere) = E*(4*pi*r_b^2) .

The right hand side is simply, the charge density times the volume of the arbitrary sphere whose r < r_b.
Q = (4/3 pi r^3) * rho.

Therefore the resulting ratio should go as

E(r < r_b) = r^3 * rho /(3 * r_b^2 *epsilon).

Check with the instructor. The computer's solution may have a typo.
 
  • #3
Seraph404 said:
[tex]\int[/tex] EA = Q/[tex]\epsilon[/tex]0

The hint it gave me was that my answer should not depend on rb. Why wouldn't it?

Hi Seraph404! :smile:

(have a pi: π and a rho: ρ :wink:)

Hint: E is the same for all points on the sphere of radius r < rb

so what is the flux through the surface of that sphere?
 
  • #4
Ah.. ><
I'm almost 99% sure that the problem is me. But I'll ask anyway if nobody else on here can spot the mistake.
 
  • #5
tiny-tim said:
Hi Seraph404! :smile:

(have a pi: π and a rho: ρ :wink:)

Hint: E is the same for all points on the sphere of radius r < rb

so what is the flux through the surface of that sphere?

Um. I guess radially outward? But how does that remove my need for rb in the expression?

I should also have mentioned that I'm struggling with the material. My professor is absolutely no help. I've started skipping lecture just to hang out in the Physics helpdesk and work problems, since that's the only way I'm going to learn any of this.
 
  • #6
Ok, I did the math on paper instead of in my head.

Here is the answer. Got a pen and paper? Nope that pen is dry. Try another one.
Ahh one that works.

Left hand side = E * (area of a sphere ENCLOSING THE CHARGE)
= E * (4*pi*r^2)

Right hand side = Q/epsilon
= (volume of ENCLOSED CHARGE) * (Charge density)
= (4/3) * pi * r^3 * rho / epsilon

Result => E = (r * rho) / (3 * epsilon)

You can check this at the boundary where r = r_b. Let me show you.
 
  • #7
Hi Mr.Amin! Welcome to PF! :smile:

But please help by giving just hints on this forum, rather than by actually doing. :wink:

Seraph404 said:
Um. I guess radially outward? But how does that remove my need for rb in the expression?

Yes, but how much is it? E(r) times what?

And then what should that flux be equal to?

(what is flux through a surface always equal to? :wink:)
 
  • #8
The continues from my previous post, I can't type as fast as you!

r > r_b

Left hand side = E * (4 * pi * r^2)

Right hand side = (4/3) * pi * r^2 * rho / epsilon

Solve for E

E = (r^3 * rho) / (3 * epsilon * r^2)
= (r * rho) / (3 * epsilon).

You could have gotten here if you used E = 1/(4*pi*epsilon) * Q/r^2 .

Now check E(r = r_b) for both equations.

for r < r_b
E(r = r_b) = (r_b * rho) / (3 * epsilon)

for r > r_b
E(r = r_b) = (r_b * rho) / (3 * epsilon).

The reason why r_b does not come into your previous question is because a uniform distribution of charge creates essentially no net electric field. If you pretend your sensor has penetrated into the charged sphere, everything at the radius above it will add to zero field. ONLY the charge in the sphere below the probed radius will generate the non-zero electric field.

This works for gravity too since Gauss's law works for g-fields as well.
 
  • #9
Oh.. I guess I didn't realize that the area in Gauss's law was for enclosed charge only. Anyway,Thanks guys! I will definitely refer back to this post when I work future problems! Hopefully, the more of these I work, the better my understanding of this subject will be.

My main problem is that our prof began by throwing equations at us (not even bothering to derive them) and scribbling down example problems here and there that hardly any of us could make out. So conceptually, I have little to no idea what is going on in this subject. The book is a big help, but sometimes I need things drawn out for me before I'll really understand. Anyway, thanks again!
 

1. What is Gauss's Law?

Gauss's Law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface.

2. What is a "Basic Gauss's Law problem"?

A Basic Gauss's Law problem is a simplified version of a more complex problem, typically involving a uniform electric field and a simple geometric shape, used to demonstrate the application of Gauss's Law.

3. How do you determine the electric flux through a closed surface?

The electric flux through a closed surface can be determined by calculating the dot product of the electric field and the surface area vector at each point on the surface and then summing these values over the entire surface.

4. What is the equation for Gauss's Law?

The equation for Gauss's Law is ΦE = Qenc0, where ΦE is the electric flux, Qenc is the enclosed charge, and ε0 is the permittivity of free space.

5. How is Gauss's Law used in real-world applications?

Gauss's Law is used in many real-world applications, such as calculating the electric field inside a charged conducting sphere, designing capacitors, and determining the electric field of a point charge or a charged line or plane. It is also a key component in understanding and analyzing electromagnetic waves.

Similar threads

  • Introductory Physics Homework Help
Replies
26
Views
554
  • Introductory Physics Homework Help
Replies
5
Views
203
  • Introductory Physics Homework Help
Replies
1
Views
808
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
4K
  • Introductory Physics Homework Help
Replies
6
Views
820
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top