Calculating the electric potential

[physicsisfun0]

Homework Statement

We have the cross section of a metal pipe that has been split into four sections. Three of the sections have a constant electric potential, V_o. The fourth section is grounded so electric potential is zero. We are looking for electric potential inside and outside of the pipe.

Homework Equations

For electric potential I have:
V = a_o + b_oln(s) +Σ(s^v(a_vcos(vφ) + b_osin(vφ)) + s^-v(c_vcos(vφ) + d_vsin(vφ))
We also know boundary conditions:

The Attempt at a Solution

(I think this next part is right)
And electric potential inside becomes:
V = a_o + b_oln(s) +Σ(s^v(a_vcos(vφ) + b_osin(vφ))
And electric potential outside becomes:
V = a_o + b_oln(s) +Σs^-v(c_vcos(vφ) + d_vsin(vφ))

There is also symmetry along the x-axis so we can ignore the sin(vφ) contribution:
V = a_o + b_oln(s) +Σ(s^v(a_vcos(vφ))
V = a_o + b_oln(s) +Σ(s^-v(a_vcos(vφ))

Do I need to solve for a_o and b_o? How would I use the boundary conditions?

 

[TSny]

Welcome to PF!

Note that the first equation above implies that ##7 \pi / 4## is less than ##\pi/4##. But, the intention is clear.

V = a_o + b_oln(s) +Σ(s^v(a_vcos(vφ))
V = a_o + b_oln(s) +Σ(s^-v(a_vcos(vφ))

OK, but it might be confusing to use the same notation for the coefficients for the two different regions.

Do I need to solve for a_o and b_o? How would I use the boundary conditions?

For the region inside the pipe, note that s can be zero. What does that tell you about the value of b_o for the inside region?
For the region outside the pipe, note that s can become arbitrarily large. What does that tell you about b_o for the outside region?

To find a_o, set s equal to the radius of the pipe in the above two equations. For each equation, integrate both sides of the equation with respect to φ from 0 to 2π. The boundary condition tells you V as a function of φ for the integration of the left side.

To find other coefficients a_n, multiply both sides of the equation by cos(nφ) before integrating.