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Determining net force using Gauss' Law

  • #1

Homework Statement


We have a uniformly charged, non-conducting sphere (charge per volume,ρ, and radius, R). Then a uniformly charged ruler (charge per length,λ, and length, d) is aligned radially almost touching the surface of the sphere. Determine the net force experienced by the ruler.

Homework Equations


∫E⋅dA=Qenclosed0
I think thats all...

The Attempt at a Solution


I drew my gaussian surface like Gaussian surface.jpg
∫E⋅dA=Qenclosed0
ρ=(Q/V)=Q/(4/3)πR2→Q=ρ(4/3)πR2
A=4πR12
∫E⋅dA=Qenclosed0→E4πR12=ρ(4/3)πR20
E=ρR2/3ε0R12

Now I have the electric field of the sphere at its surface
When I place a ruler almost touching the sphere aligned radially im not sure what to do... The ruler also has an electric field which we could calculate using gauss's law (the shape would be a box?). I know the spheres electric field interacts with the charged ruler...
Any help would be appreciated!
 

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Answers and Replies

  • #2
TSny
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→Q=ρ(4/3)πR2
Check if you have the correct power on the R.

E=ρR2/3ε0R12

Now I have the electric field of the sphere at its surface
OK, except for the power on R. Does this represent the field at the surface of the sphere or at a point a distance R1 from the center of the sphere (where R1 > R)?

When I place a ruler almost touching the sphere aligned radially im not sure what to do... The ruler also has an electric field which we could calculate using gauss's law (the shape would be a box?). I know the spheres electric field interacts with the charged ruler...
Any help would be appreciated!
The net force on the ruler is due to the field of the sphere. If you took an infinitesimal segment of the ruler of length dr located a distance r from the center of the sphere, how could you express the electric force on this segment due to the field of the sphere?
 
  • #3
Check if you have the correct power on the R.
So does is the density of the sphere: ρ=Q/(4/3)πR3? "R" being the radius of the sphere not gaussian surface? Just realized I made a mistake with volume earlier.
The Qenclosed in gauss's law is related to the charge density of the sphere, correct?
The net force on the ruler is due to the field of the sphere. If you took an infinitesimal segment of the ruler of length dr located a distance r from the center of the sphere, how could you express the electric force on this segment due to the field of the sphere?
So I understand what you are saying, but not sure I can remember how to do that.
The field would be changing with every small segment of ruler, there is clearly an integral involved whose limits would be r→r+L (Assuming the ruler's length is L).
I know this is wrong but:
F=q∫E, Limits stated above...
 
  • #4
TSny
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So does is the density of the sphere: ρ=Q/(4/3)πR3? "R" being the radius of the sphere not gaussian surface?
Yes. However, you should enclose the entire denominator in parentheses (or brackets) to make it clear what's in the denominator. Thus,
ρ=Q/[(4/3)πR3]

The Qenclosed in gauss's law is related to the charge density of the sphere, correct?
You could say that. It's clear that the charge enclosed in your chosen Gaussian surface is the entire charge Q of the sphere. You can express Q in terms of ρ. Your derivation of E in your first post looks fine to me except for the power of R.

So I understand what you are saying, but not sure I can remember how to do that.
The field would be changing with every small segment of ruler, there is clearly an integral involved whose limits would be r→r+L (Assuming the ruler's length is L).
I know this is wrong but:
F=q∫E, Limits stated above...
Yes, you will need to set up an integral. The total force F on the stick is the sum of the forces dF on the individual infinitesimal elements of the stick:

F = ∫dF ##\,\,\,\,##So, you need an explicit expression for dF.

An infinitesimal element of the stick will have an infinitesimal length dr along the radial direction and it will be located at some distance r from the center of the sphere. If E represents the value of the field of the sphere at this location, how can you express the force dF on this element in terms of dq and E. Then think about how to express dq in terms of dr and how to express E in terms of r.
 
  • #5
Yes, you will need to set up an integral. The total force F on the stick is the sum of the forces dF on the individual infinitesimal elements of the stick:

F = ∫dF \,\,\,\,So, you need an explicit expression for dF.

An infinitesimal element of the stick will have an infinitesimal length dr along the radial direction and it will be located at some distance r from the center of the sphere. If E represents the value of the field of the sphere at this location, how can you express the force dF on this element in terms of dq and E. Then think about how to express dq in terms of dr and how to express E in terms of r.
I could say

dF=Edq=EσdA

dA=dr x the width of ruler?

Im not sure...
 
  • #6
TSny
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I could say

dF=Edq=EσdA

dA=dr x the width of ruler?

Im not sure...
You can treat the ruler as a thin rod with linear charge density λ. Express dq in terms of λ.
 
  • #7
You can treat the ruler as a thin rod with linear charge density λ. Express dq in terms of λ.
something like:
dF=Eλdr, then write E=kQ/r
F=∫dF=∫(kQ/r)λdr (limits from R being the radius of the sphere to R+L)?
 
  • #8
TSny
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something like:
dF=Eλdr, then write E=kQ/r
F=∫dF=∫(kQ/r)λdr (limits from R being the radius of the sphere to R+L)?
Yes. You should express your final result in terms of the given parameters: ρ, λ, R, and d.
 
  • #9
rude man
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something like:
dF=Eλdr, then write E=kQ/r
I think your denominator is incorrect. Look again at the remark at the beginning of post 2 ...
 

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