Cylindrical Capacitor

Question

http://crab.rutgers.edu/~gurfink/physics2/practice_exams/node1.html

Need help w/ problem 5. thanks in advance

Answer 1

Use Gauss's Law to determine the electric fields in the different regions.

The area integral of the electric field over any closed surface is equal to the net charge enclosed in the surface divided by the permittivity of space.

Then to get the voltage

V = E d

where d is the separation

Answer 2

The trick is that inside a metal volume, the field is always zero.  

And outside in cylindrical symmetry, E = (Q/L) / (2*pi*e0*r)

where Q is total enclosed charge (taking into account all surfaces).

So first step is to use the above to give field from a to b.  Note it is zero for r<a (inside conductor).

Next, you know the field must be zero between b and c (again, inside conductor).  From that you could figure out that there must be -Q on the inner surface, -2Q on outer surface.  Finally, outside c, there is only a net charge of -2Q, so you have same equation, except with -2Q instead of Q.

Now, for potential, you take negative integral of E with respect to r moving from b.  Note that it is largest at r<=a, decreases to b, constant through c, then falls off to zero moving away from c.  

Note that you will get some natural logs in this integral.

Capacitance is charge of divided by voltage -- or charge on a, divided by voltage of a.

Energy is given by 1/2 CV^2 (again, voltage on center conductor)