Electric Potential: Q, a, b, Vb & Va

[brad sue]

Hi,
I would like to have some guidance with this problem. i would like to know please, if I do good or bad.

A coaxial line is made of two perfect conductors. The inner cylinder has radius a and the radius of the outer one is b. The outer conductor is grounded.
Suppose that the inner conductor carries 1[C] of charge per meter of its length, that is its equivalent line charge density is ρlo=1[C/m].

a- Find the potential difference between the two conductors.
b- What is the potential of the outer conductor?
c- What is the potential of the inner conductor?

a-
V= (Q/4*pi* ε0)( 1/a-1/b)

with Q=?

b-
Vb=0[V]

c-

Va=(Q/4*pi*ε0)(1/a)

I have some issue with the value of Q. since the length is not specified should Q be Q=1*L=L (Length)??

 

[interested_learner]

The trick here is to figure out the charge density which is :

$$\rho = \frac {1 Coulomb} {2 \pi a meter^2}$$

You have to divide the C/meter by the distance around which of course is a $$2 a \pi$$ meter. That gives you the charge density on the surface.

Given the density, you can solve for the electrical field which is what you were close to doing in your answer in part a.

That isn't quite right since the electric field has a direction, but that can be ignored in this problem if we assume everything is along the radial direction.

The only problem is that isn't the potential. You need to integrate the negative the distance from b to a to get the potential.

 

[Alch]

Does anyone know what the correct solutions are to this problem?