"Potential of Concentric Cylindrical Insulator and Conducting Shell"

In summary, a charged ring has an electric field that is inversely proportional to the distance from the ring.
  • #1
tylerlu94
4
0

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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  • #2
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
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
 
  • #4
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.
 
  • #5
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!
 
  • #6
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
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?
 
  • #8
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?
 

FAQ: "Potential of Concentric Cylindrical Insulator and Conducting Shell"

What is the purpose of using a concentric cylindrical insulator and conducting shell?

The purpose of using a concentric cylindrical insulator and conducting shell is to create a capacitor, which is a device used to store electrical energy. The insulator, which is typically made of a non-conductive material like rubber or plastic, separates the two conductive shells and prevents the flow of electricity. The conducting shell, which is usually made of metal, is used to carry the electric charge.

How does the shape of the concentric cylindrical insulator and conducting shell affect its potential?

The shape of the concentric cylindrical insulator and conducting shell affects its potential by increasing the capacitance, or the ability to store electric charge. The larger the surface area of the conducting shell, the higher the potential will be. Additionally, the distance between the two shells also plays a role in the potential, as a smaller distance will result in a higher potential.

Can the potential of a concentric cylindrical insulator and conducting shell be changed?

Yes, the potential of a concentric cylindrical insulator and conducting shell can be changed by altering the size of the insulator, the surface area of the conducting shell, or the distance between the two shells. Additionally, the type of material used for the insulator and conducting shell can also affect the potential.

How does the potential of a concentric cylindrical insulator and conducting shell compare to other types of capacitors?

The potential of a concentric cylindrical insulator and conducting shell is dependent on the size and shape of the components, so it can vary. However, in general, this type of capacitor has a higher potential compared to other types, such as parallel plate capacitors, due to the increased surface area and smaller distance between the shells.

Are there any limitations to using a concentric cylindrical insulator and conducting shell?

One limitation of using a concentric cylindrical insulator and conducting shell is that it can only store a certain amount of electrical energy before reaching its maximum potential. Additionally, the size and shape of the components may be limited by practical considerations, making it difficult to create very large or small capacitors. Lastly, the insulator material must have a high breakdown voltage to prevent electric discharge and maintain the integrity of the capacitor.

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