Coaxial pair of infinitely long charged solid conductors

[ness9660]

Consider a coaxial pair of infinitely long charged solid conductors. The inner conductor has a radius R, while the outer conductor has an inner radius 2R and an outer radius 3R.
The inner conductor has a linear charge density 2λ, while the outer conductor has a net linear charge density of -3λ.

A) Using Gauss’ law and the properties of conductors, what are the linear charge densities on the inner and outer surfaces of the outer conductor.
B) Use Gauss’ law to find an expression for the electric field as a function of radius for all four regions.

A) By definition linear charge density is Q/L. So for the inner conductor with 2λ I want to say it is 2Q/infinity but this cannot be right. I am sure using Guass's law produces a correct answer but I cannot see anyway to relate it to λ or length of the conductors for that matter

B) For this part I would assume you have to setup two integrals, one for the field produced by the inner conductor and for the outer conductor? I am thinking that dA would relate to cross sectional area of the conductors and not a differential square on the surface?

 

[Andrew Mason]

A) By definition linear charge density is Q/L. So for the inner conductor with c I want to say it is 2Q/infinity but this cannot be right. I am sure using Guass's law produces a correct answer but I cannot see anyway to relate it to λ or length of the conductors for that matter

The infinite length means that there is no horizontal component to the field (horizontal components are equal and opposite). So the gaussian surface to use is a ring of width dL centred on the axis of the cable.

Place a gaussian ring surface inside the outer conductor. What is the field inside the outside conductor (inside any conductor)? What must the total enclosed charge be? Since the charge density of the inner conductor is 2λ, what must the charge density on the inner surface be?

To do the outer surface, place a gaussian ring around the whole cable. What is the total enclosed charge? That and the previous answer should enable you to find the charge density on the outer surface.

B) For this part I would assume you have to setup two integrals, one for the field produced by the inner conductor and for the outer conductor? I am thinking that dA would relate to cross sectional area of the conductors and not a differential square on the surface?

Just apply Gauss' law: a) inside the inner conductor b) between the inner and outer conductor, c) inside the outer conductor and d) outside the outer conductor. It would help to plot the field on a graph as a function of r.

AM

 

[kyle8921]

I would approach this problem by first identifying the key information and variables given in the scenario. We have two infinitely long charged solid conductors, one with a radius R and the other with inner and outer radii of 2R and 3R respectively. The inner conductor has a linear charge density of 2λ, while the outer conductor has a net linear charge density of -3λ. We are also given the use of Gauss' law, which relates the electric flux through a closed surface to the enclosed charge.

A) To determine the linear charge densities on the inner and outer surfaces of the outer conductor, we can use Gauss' law. This law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε0). In this case, we can consider a cylindrical Gaussian surface with a radius r, centered between the two conductors. Since the conductors are infinitely long, we can ignore the effects of the ends of the cylinders.

For the inner conductor, the charge enclosed by the Gaussian surface is 2λ multiplied by the length of the conductor, which is infinite. So the electric flux through the Gaussian surface is 2λ times the circumference of the circle (2πr). Setting this equal to 2λ multiplied by the permittivity of free space, we can solve for the linear charge density on the inner surface of the outer conductor.

Using the same approach for the outer conductor, the charge enclosed by the Gaussian surface is -3λ multiplied by the length of the conductor, which is also infinite. So the electric flux through the Gaussian surface is -3λ times the circumference of the circle (2πr). Setting this equal to -3λ times the permittivity of free space, we can solve for the linear charge density on the outer surface of the outer conductor.

B) To find the expression for the electric field as a function of radius for all four regions, we can again use Gauss' law. For the inner conductor, the electric field is constant and equal to 2λ/ε0. For the region between the two conductors (2R < r < 3R), the electric field is the sum of the fields produced by both conductors. This can be calculated by integrating the electric flux over the Gaussian surface, taking into account the linear charge densities on both surfaces. For the region outside the outer