A non-conducting sphere, e-field and potential.

In summary, the conversation discusses finding the electric field and potential of a non-conducting sphere with a uniform charge density inside and outside the sphere. The charge density is given by ρ = B/r for r < R and ρ = 0 for r > R, where B is a constant. The steps to finding the electric field and potential are discussed, including finding the charge of the sphere, using the equations V = -∫Edr and ∫Eda = Qin/ε, and integrating from one boundary to another. The final answers for the electric field and potential inside and outside the sphere are given as E = B/2ε and V = BR/2ε - Br/2ε, respectively.
  • #1
timnswede
101
0

Homework Statement


A non-conducting sphere of radius R has volume charge density ρ = B/r. for r<R and ρ = - for r>R. B is a constant.
a) Calculate E-field for r>R.
b) Calculate E-field for r<R.
c) Calculate potential for r>R.
d) Calculate potential for r=R.
e) Calculate potential for r<R

Homework Equations


V=-∫Edr
∫Eda=Qin/ε

The Attempt at a Solution


Well first of all I was a bit confused on the whole ρ = 0 when r>R, but apparently that just means that all of the charge is contained within the sphere.
So since it is a nonuniform charge, I first found the charge of the sphere. ∫(0 to R)ρ4πr^2dr and got 2BπR^2.
I'm pretty confident about all my answers except for part e) For part a) I did E4πr^2=2BπR^2/ε and got E=BR^2/2r^2ε. Same thing for part b, but this time the radius cancels out so B/2ε
So for part c I took the negative integral of the e-field outside of R. V=-∫(BR^2/2r^2ε)dr and ended up with BR^2/2rε. Same thing for part d, but the radius cancels out.
Now part e I did the same thing -∫B/2εdr and got -Br/2ε. I feel like it might be wrong since it's negative, and I don't really understand why the potential inside the sphere would be negative
 
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  • #2
timnswede said:

Homework Statement


A non-conducting sphere of radius R has volume charge density ρ = B/r. for r<R and ρ = - for r>R. B is a constant.
a) Calculate E-field for r>R.
b) Calculate E-field for r<R.
c) Calculate potential for r>R.
d) Calculate potential for r=R.
e) Calculate potential for r<R

Homework Equations


V=-∫Edr
∫Eda=Qin/ε

The Attempt at a Solution


Well first of all I was a bit confused on the whole ρ = 0 when r>R, but apparently that just means that all of the charge is contained within the sphere.
So since it is a nonuniform charge, I first found the charge of the sphere. ∫(0 to R)ρ4πr^2dr and got 2BπR^2.
I'm pretty confident about all my answers except for part e) For part a) I did E4πr^2=2BπR^2/ε and got E=BR^2/2r^2ε. Same thing for part b, but this time the radius cancels out so B/2ε
So for part c I took the negative integral of the e-field outside of R. V=-∫(BR^2/2r^2ε)dr and ended up with BR^2/2rε. Same thing for part d, but the radius cancels out.

Replace r=R. The radius does not cancel.

timnswede said:
Now part e I did the same thing -∫B/2εdr and got -Br/2ε. I feel like it might be wrong since it's negative, and I don't really understand why the potential inside the sphere would be negative

You get the potential with an unknown constant, C. So U = C - Br/2ε. The potential is continuous function, and you know it at R. From that, you can find C.
 
  • #3
Woops you are right, so for part d it is actually BR/2ε
Why would I integrate from R to r? What my thinking was is that I am integrating the e-field to find the potential everywhere inside the sphere, from the center, zero to the radius of the sphere R.
 
  • #4
timnswede said:
Woops you are right, so for part d it is actually BR/2ε
Why would I integrate from R to r? What my thinking was is that I am integrating the e-field to find the potential everywhere inside the sphere, from the center, zero to the radius of the sphere R.
You do not know the potential at the centre of the sphere, but you know it at r=R.

The electric field is the negative gradient of the potential. dU/dr=-E.
If you integrate it between a and b you get ## \int _a^b(dU/dr )dr = -\int_a^b Edr \rightarrow U_b-U_a= -\int_a^b Edr##
The potential was assumed zero at infinity. You do not know that it is zero at r=0 again. But it is known at r=R. Choose one boundary at R.
 
  • #5
timnswede said:
Woops you are right, so for part d it is actually BR/2ε
Why would I integrate from R to r? What my thinking was is that I am integrating the e-field to find the potential everywhere inside the sphere, from the center, zero to the radius of the sphere R.
You do not know the potential at the centre of the sphere, but you know it at r=R.

The electric field is the negative gradient of the potential. dU/dr=-E.
If you integrate it between a and b you get ## \int _a^b(dU/dr )dr = -\int_a^b Edr \rightarrow U_b-U_a= -\int_a^b Edr##
The potential was assumed zero at infinity. You do not know that it is zero at r=0 again. But it is known at r=R. Choose one boundary at R.
 
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  • #6
ehild said:
You do not know the potential at the centre of the sphere, but you know it at r=R.

The electric field is the negative gradient of the potential. dU/dr=-E.
If you integrate it between a and b you get ## \int _a^b(dU/dr )dr = -\int_a^b Edr \rightarrow U_b-U_a= -\int_a^b Edr##
The potential was assumed zero at infinity. You do not know that it is zero at r=0 again. But it is known at r=R. Choose one boundary at R.
Oh OK, that makes sense. So would it work if I did VR-Vr=-∫(from r to R)B/2ε dr. So after some integration I would get Vr=BR/ε -Br/2ε?
 
  • #7
timnswede said:
Oh OK, that makes sense. So would it work if I did VR-Vr=-∫(from r to R)B/2ε dr. So after some integration I would get Vr=BR/ε -Br/2ε?

Correct!
 
  • #8
Great, thanks! Your explanation helped a lot.
 
  • #9
You are welcome:)
 

What is a non-conducting sphere?

A non-conducting sphere is a spherical object made of a material that does not allow the flow of electric current. This means that the electrons within the material are tightly bound and cannot move freely.

What is an e-field?

An e-field, or electric field, is a property of space that describes the influence that a charged object has on other charged objects in its vicinity. It is represented by lines that indicate the direction and strength of the force that would be experienced by a positive charge placed in that field.

What is potential in relation to a non-conducting sphere?

Potential, or electric potential, is a measure of the amount of electrical potential energy that a charged object has. In the case of a non-conducting sphere, the potential is determined by the distribution of charge on the surface of the sphere and the distance from the center of the sphere.

How is potential related to the e-field of a non-conducting sphere?

The potential of a non-conducting sphere is directly related to the e-field surrounding it. The direction of the e-field points from areas of high potential to low potential. In the case of a non-conducting sphere, the e-field will be strongest near the surface of the sphere where the potential is highest.

How does the e-field and potential change with distance from a non-conducting sphere?

The e-field and potential both decrease with distance from a non-conducting sphere. The e-field follows the inverse square law, meaning that it decreases by the square of the distance from the sphere. The potential also decreases with distance, but at a slower rate, following the inverse distance law.

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