"Potential of Concentric Cylindrical Insulator and Conducting Shell"

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SUMMARY

The discussion focuses on calculating the potential difference between an infinitely long solid insulating cylinder and a concentric conducting shell. The insulating cylinder has a charge density of 30 μC/m³, while the conducting shell has a linear charge density of -0.32 μC/m. The electric field equations used include E = λ/(2πε₀r) and ΔV = -∫E*dr. Participants clarify the correct approach to finding the electric field and potential difference, emphasizing the need to consider both the insulating cylinder and the conducting shell in the calculations.

PREREQUISITES
  • Understanding of electrostatics, specifically Gauss's Law
  • Familiarity with electric field calculations for cylindrical charge distributions
  • Knowledge of potential difference and its mathematical representation
  • Basic calculus, including integration techniques
NEXT STEPS
  • Study the application of Gauss's Law for cylindrical symmetry
  • Learn about electric fields generated by line charges
  • Explore potential difference calculations in electrostatics
  • Review integration techniques for solving electric field problems
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone involved in solving electrostatic problems related to cylindrical charge distributions.

tylerlu94
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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 ??
 
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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##.
 
Zondrina said:
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 learned that in math yet. I only know integral and derivatives.. Can you please explain it in a easier way? Thank you
 
tylerlu94 said:
Hmmmm Sorry I am a little bit confused. What's differential charge element?? I guess haven't learned 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.
 
Zondrina said:
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!
 
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.
 
Zondrina said:
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?
 
tylerlu94 said:
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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