Concentric Cylindrical Conducting Shells Potential Difference

Click For Summary
SUMMARY

The discussion focuses on calculating the potential difference between two concentric cylindrical conducting shells. The inner shell has a radius of 4.5 cm and a linear charge density of -0.35 μC/m, while the outer shell has an inner radius of 17.1 cm, an outer radius of 20.1 cm, and a linear charge density of 0.35 μC/m. The potential difference is determined using the formula ΔV = V_b - V_a = -∫ E · dℓ, where the electric field E is derived from Gauss's law. The user expresses uncertainty about calculating the electric field for a three-dimensional object given only the linear charge density.

PREREQUISITES
  • Understanding of Gauss's Law for electric fields
  • Familiarity with cylindrical coordinates in electrostatics
  • Knowledge of electric potential and potential difference concepts
  • Basic calculus for evaluating integrals
NEXT STEPS
  • Study Gauss's Law applications for cylindrical symmetry
  • Learn how to derive electric fields from linear charge densities
  • Explore the concept of electric potential in electrostatics
  • Practice solving integrals involving electric fields and potential differences
USEFUL FOR

Students studying electromagnetism, particularly those tackling problems involving electric fields and potential differences in cylindrical geometries.

jeff24
Messages
1
Reaction score
0

Homework Statement



An infiinitely long solid conducting cylindrical shell of radius a = 4.5 cm and negligible thickness is positioned with its symmetry axis along the z-axis as shown. The shell is charged, having a linear charge density λinner = -0.35 μC/m. Concentric with the shell is another cylindrical conducting shell of inner radius b = 17.1 cm, and outer radius c = 20.1 cm. This conducting shell has a linear charge density λ outer = 0.35μC/m.

https://www.physicsbrain.com/images/content/EM/08/h8_cylinder.png

What is V(c) – V(a), the potential difference between the the two cylindrical shells?

Homework Equations



[tex]\Delta V = V_b - V_a = \frac{{\Delta U}}{{q_0 }} = - \int_a^b {E \cdot d\ell }[/tex]

V(c)-V(a) = -integral from A to C of the Electric Field dot dl

The Attempt at a Solution



I don't know where to start. I don't even know how to get the electric field or charge of a 3-D object (outer cylindrical shell) when I'm only given lambda, charge/meter.
 
Last edited by a moderator:
Physics news on Phys.org
Use Gauss's law to get the electric field.
 

Similar threads

Replies
5
Views
1K
Replies
2
Views
3K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
7
Views
7K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
5
Views
49K