"Potential of Concentric Cylindrical Insulator and Conducting Shell"

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Homework Statement


h6_cylinder.png

An infinitely long solid insulating cylinder of radius a = 2.5 cm is positioned with its symmetry axis along the z-axis as shown. The cylinder is uniformly charged with a charge density ρ = 30 μC/m3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 15.3 cm, and outer radius c = 19.3 cm. The conducting shell has a linear charge density λ = -0.32μC/m.d=51cm.

What is V(c) - V(a), the potentital difference between the outer surface of the conductor and the outer surface of the insulator?

Homework Equations



ΔV=-∫E*dr.
E=λ/2ε0

The Attempt at a Solution



ΔV=Vab+Vbc; Eab=λ/2ε0;
Ebc I used flux EA=Qenc/ε0 → E=(Qenc)/2*∏*r*L*ε0 → Vbc=∫(b→c) (Qenc)/(2*∏*r*L*ε0 ) *dr
This is basically how I did it, But I got the wrong answer -4397..Please help me ??
 

Answers and Replies

  • #2
STEMucator
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What is the potential of the outer charge distribution? Take a differential charge element ##dq = \lambda ds## to find it.

How about the inner one now? Take a differential charge element ##dq = \rho dV##.
 
  • #3
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What is the potential of the outer charge distribution? Take a differential charge element ##dq = \lambda ds## to find it.

How about the inner one now? Take a differential charge element ##dq = \rho dV##.

Hmmmm Sorry I am a little bit confused. What's differential charge element?? I guess haven't learnt that in math yet. I only know integral and derivatives.. Can you please explain it in a easier way? Thank you
 
  • #4
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Hmmmm Sorry I am a little bit confused. What's differential charge element?? I guess haven't learnt that in math yet. I only know integral and derivatives.. Can you please explain it in a easier way? Thank you
Wow sorry ignore my last post. I should have read your post more carefully.

What did you compute for the electric field of the inner cylinder? Your formula looks wrong it should be:

##E = \frac{\lambda}{2 \pi \epsilon_0 r}##

I was a bit thrown off by that random point (d,d) I saw in the image.
 
  • #5
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Wow sorry ignore my last post. I should have read your post more carefully.

What did you compute for the electric field of the inner cylinder? Your formula looks wrong it should be:

##E = \frac{\lambda}{2 \pi \epsilon_0 r}##

I was a bit thrown off by that random point (d,d) I saw in the image.
hmmm sorry I can't remember. But I think the basic idea of my solution was wrong.. because I basically just made it up... Can you please let me know how would the correct solution be?(or just the basic idea and structure of it )
Thank you!
 
  • #6
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Find the electric field of the infinitely long line of charge at a radial distance ##r## away.

Integrate this uniform field over the path length.
 
  • #7
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Find the electric field of the infinitely long line of charge at a radial distance ##r## away.

Integrate this uniform field over the path length.
Just the infinite line? So ignore the field by the shell? but the r starts from the outer surface of the shell though?
 
  • #8
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Just the infinite line? So ignore the field by the shell? but the r starts from the outer surface of the shell though?
No don't ignore the field of the shell.

What is the electric field due to a charged ring?
 

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