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Gauss's Law on cylinders

  • Thread starter ndoc
  • Start date
  • #1
13
0

Homework Statement


The figure below shows a portion of an infinitely long, concentric cable in cross section. The inner conductor has a linear charge density of λ = 6.10 nC/m and the outer conductor has no net charge.

http://www.webassign.net/tipler/23-36alt.gif

(a) Find the electric field for all values of R, where R is the perpendicular distance from the common axis of the cylindrical system. (Use R as necessary.)
1.50 cm R 4.50 cm
R 6.50 cm

(b) What are the surface charge densities on the inside and the outside surfaces of the outer conductor?

Homework Equations



∫EdA = Qinside/ɛo



The Attempt at a Solution



Using the above equation I would get E(2pi*r^2*h) = Qenclosed/ɛo, but it is infinite in length and therefore no height is given. Any help appreciated.
 

Answers and Replies

  • #2
783
9
Then you would just have a surface charge per unit length of the infinitely long conductor. That is the beauty of the infinitely long objects.

Thanks
Matt
 
  • #3
50
0
Hi ndoc,

Surface density (λ) = Total charge (Q)/Length.

If you set Gauss's law equivalent to a general form of Coulomb's law, EA=Q/ɛo, you will find that the cylinder's height, or length, does not have any significance.

i.e.,

E*[2pi(r^2)*h]= (λ*h)/ɛo =>

E=λ/[2piɛo*(r^2)]
 

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