Potential inbetween two coaxial cylinders

[aftershock]

Homework Statement

There are 2 coaxial cylindrical conductors. The inner cylinder has radius a, while the outer cylinder has radius b. There is no charge in the region a < r < b. If the inner cylinder is at potential V_o and the outer cylinder is grounded, we want to find the potential in the region between the cylinders. We assume L >> b > a and neglect end effects.

a) Write Laplace's equation in cylindrical coordinates.

b) Assuming V(r) is a function of the axial distance r alone, integrate the differential equation and use the boundary conditions to find V(r) , a ≤ r ≤ b.

Homework Equations

∇²V = 0

The Attempt at a Solution

Laplace's equation in cylindrical coordinates would be

(1/r)(d/dr)(r*dV/dr) =0

Since there is no phi or z dependence. Also I know it's a partial and not d but I can't type the symbol.

To solve this

(1/r)(d/dr)(r*dV/dr) =0

(d/dr)(r*dV/dr) =0

r*dV/dr = C

dV/dr = C/r

dV = C/r dr

V = C*ln(r) + K , where C and K are constants.

Using the boundary conditions:

0 = C*ln(b) + K

V_o = C*ln(a) + K

and using that to solve for constants:

V = V_o/[ln(a/b)] * ln(r/b)

Is that right?

 

[mfb]

The result is correct and the other parts look fine, too.

LaTeX code can be used to write partial derivatives: $[/color]\frac{\partial}{\partial r}$[/color] --> $\frac{\partial}{\partial r}$.

 

[rude man]

Homework Statement

V = V_o/[ln(a/b)] * ln(r/b)

Is that right?

Confirmed using Gauss' theorem.