Charge enclosed in a sphere

  • #1
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Homework Statement



A solid insulating sphere of radius R is charged so that the charge density is proportional to the distance to the center, p(r)=kr, where k is a constant.

Find Q(r), the charge enclosed by a concentric spherical surface of radius r<= R. What is the total charge of the sphere?

Homework Equations





The Attempt at a Solution



I thought that Q=[Int from 0 to r] kr dV, and in spherical coordinates for radius, dV = r^2 dr.

So I have Q=[int from 0 to r] kr^3 dr = k [int, 0, r] r^3 dr = kr^4 / 4
 

Answers and Replies

  • #2
ehild
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The Attempt at a Solution



I thought that Q=[Int from 0 to r] kr dV, and in spherical coordinates for radius, dV = r^2 dr.

So I have Q=[int from 0 to r] kr^3 dr = k [int, 0, r] r^3 dr = kr^4 / 4
The volume element is a shell of radius r and thickness dr, which volume is dV=4pi r2dr

ehild
 
  • #3
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The volume element is a shell of radius r and thickness dr, which volume is dV=4pi r2dr

ehild
then when do I use r^2 sin(theta) dr d(theta) d(phi)?
 
  • #4
ehild
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then when do I use r^2 sin(theta) dr d(theta) d(phi)?
You are right now, the volume element is dV=r^2sin(theta)dr d(theta) d(phi) in spherical coordinates, but not r^2dr as you wrote in your first post. Integrate the charge density for the sphere of radius r.

ehild
 
  • #5
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So lets say a question gives you the charge density p=kr where k is a constant, and it wants you to find the electric field inside a solid sphere of radius R.

Id say that [int] E ds = q/e0

and i can pull out E, -> E [int] ds = q/e0

and using what you said, E [int] ds = [int] kr r^2sin(theta)dr d(theta) d(phi) / e0

I would be left with 4piR^2 E = (4piR^3)/3e0 -> E=R / 3e0 ?
 
  • #6
ehild
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So lets say a question gives you the charge density p=kr where k is a constant, and it wants you to find the electric field inside a solid sphere of radius R.

Id say that [int] E ds = q/e0

and i can pull out E, -> E [int] ds = q/e0

and using what you said, E [int] ds = [int] kr r^2sin(theta)dr d(theta) d(phi) / e0

I would be left with 4piR^2 E = (4piR^3)/3e0 -> E=R / 3e0 ?
You have to integrate kr^3 with respect to r.

ehild
 
  • #7
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So would the total enclosed charge be [int 0,R] kr * 4pi r^2 dr = k pi R^4 ?

and then I would get (k pi R^4) / (4 pi R^2 e0) = E = kR/4e0
 
Last edited:
  • #8
ehild
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So would the total enclosed charge be [int 0,R] kr * 4pi r^2 dr = k pi R^4 ?
Yes.

ehild
 
  • #9
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Yes.

ehild
So if i use [int] E ds = k pi R^4 / e0, would I get E = kR^4 / 4r^2 e0 ?
 
Last edited:
  • #10
ehild
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So if i use [int] E ds = k pi R^4 / e0, would I get E = kR^4 / 4r^2 e0 ?
It depends what you mean on R and r and [int] E ds.

ehild
 
  • #11
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R is the radius of the sphere, r is your variable radius

it'd be the surface integral of E ds = k pi R^4 / e0

and the surface would be 4 pi r^2, divide on both sides leaves you with E = kR^4 / 4r^2 e0
 
  • #12
ehild
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Gauss' Law states that the surface integral of the electric field is equal to the enclosed charge divided by ε0. If you want to find the electric field at distance r from the centre of the charged sphere of radius R, you calculate the surface integral of E for a sphere of radius r. If r>R it encloses the whole charge, and your formula is rigth. If r≤R the enclosed charge is the charge inside the sphere of radius r. In this case, E=kr^4/(4r^2 ε0)=kr^2(4 ε0).

ehild
 

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