Calculating Surface Charge Densities in Concentric Cables

[nns91]

Homework Statement

1. A portion of an infinitely long, concentric cable. The inner conductor carries a charge of 6 nC/m; the outer conductor is uncharged. The radius of the inner conductor is 1.5 cm, of the outer conductor is 6.5cm and of the inner surface of the outer conductor is 4.5 cm.

Find the surface charge densities on the inside and the outside surface of the outer conductor

Homework Equations

density= Q/A

The Attempt at a Solution

1. So I have found that the Electric Field in 1.5<r<4.5 and r>6.5 is E= 108/r (N/C) and Electric Field in 0<r<1.5 and 4.5<r<6.5 is E=0.

So area of the cylinder is 2*pi*L. Thus

Density= Q/A = Q/(2*pi*r*L). However, since this cable is infinitely long, how do I get L then ?

 

[Andrew Mason]

Homework Statement

1. A portion of an infinitely long, concentric cable. The inner conductor carries a charge of 6 nC/m; the outer conductor is uncharged. The radius of the inner conductor is 1.5 cm, of the outer conductor is 6.5cm and of the inner surface of the outer conductor is 4.5 cm.

Find the surface charge densities on the inside and the outside surface of the outer conductor

Homework Equations

density= Q/A

The Attempt at a Solution

1. So I have found that the Electric Field in 1.5<r<4.5 and r>6.5 is E= 108/r (N/C) and Electric Field in 0<r<1.5 and 4.5<r<6.5 is E=0.

So area of the cylinder is 2*pi*L. Thus

Density= Q/A = Q/(2*pi*r*L). However, since this cable is infinitely long, how do I get L then ?

The "infinitely long" cable means the field lines have no component along the length of the cable ie. they are all perpendicular to the long axis of the cable.

This is a Gauss' law problem. Imagine a Gaussian ring of width dL with the outer surface in the middle of the outer conductor (ie. where the field must be 0), and with the inner surface inside the inner ring (ie where the field is also 0). Since there are no lateral components to the field, you only have to concern yourself with the inner and outer surfaces of this ring.

[tex]\int \vec E \cdot dA = \frac{q_{encl}}{\epsilon_0}[/tex]

What is the LS of this equation? What is the enclosed charge? From that, you can work out what the charge density must be on the inner surface of the outer conductor (you know the charge density of the inner conductor so you can work out what the charge density on the outer conductor must be in order to achieve the total enclosed charge).

AM

 

[kerimek]

I would approach this problem by first understanding the concept of surface charge density and its relationship to electric fields. Surface charge density is defined as the amount of charge per unit area on a surface. In this problem, we are given the charge on the inner conductor, and we need to find the surface charge densities on the inside and outside surfaces of the outer conductor.

To solve this problem, we can use the formula for surface charge density, which is Q/A, where Q is the charge and A is the area. In this case, we need to find the area of the inner and outer surfaces of the outer conductor.

For the inside surface, we can use the formula for the area of a cylinder, which is 2*pi*r*L, where r is the radius and L is the length. However, since the cable is infinitely long, we can assume that the length is also infinite. Therefore, the area of the inner surface is simply 2*pi*r.

For the outside surface, we can use the same formula, but this time the radius will be 4.5 cm, since that is the inner surface of the outer conductor. Therefore, the area of the outer surface is 2*pi*4.5 = 9*pi cm^2.

Now, we can plug these values into the formula for surface charge density. For the inside surface, we have Q = 6 nC/m and A = 2*pi*1.5 cm^2. This gives us a surface charge density of 2 nC/m^2.

For the outside surface, we have Q = 0 (since the outer conductor is uncharged) and A = 9*pi cm^2. This gives us a surface charge density of 0 nC/m^2.

In conclusion, the surface charge density on the inside surface of the outer conductor is 2 nC/m^2 and the surface charge density on the outside surface is 0 nC/m^2. This shows that the electric field inside the cable is non-zero, while the electric field outside the cable is zero. This makes sense, since the outer conductor is uncharged and therefore does not contribute to the electric field outside the cable.