Calculating the electric potential

  • #1
physicsisfun0
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Homework Statement


upload_2018-11-13_21-27-41.png

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, Vo. 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 = ao + boln(s) +Σ(sv(avcos(vφ) + bosin(vφ)) + s-v(cvcos(vφ) + dvsin(vφ))
We also know boundary conditions:
upload_2018-11-13_21-40-10.png

The Attempt at a Solution


(I think this next part is right)
And electric potential inside becomes:
V = ao + boln(s) +Σ(sv(avcos(vφ) + bosin(vφ))
And electric potential outside becomes:
V = ao + boln(s) +Σs-v(cvcos(vφ) + dvsin(vφ))

There is also symmetry along the x-axis so we can ignore the sin(vφ) contribution:
V = ao + boln(s) +Σ(sv(avcos(vφ))
V = ao + boln(s) +Σ(s-v(avcos(vφ))

Do I need to solve for ao and bo? How would I use the boundary conditions?
 

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Answers and Replies

  • #2
TSny
Homework Helper
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Welcome to PF!

upload_2018-11-14_22-5-13.png
Note that the first equation above implies that ##7 \pi / 4## is less than ##\pi/4##. But, the intention is clear.

V = ao + boln(s) +Σ(sv(avcos(vφ))
V = ao + boln(s) +Σ(s-v(avcos(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 ao and bo? 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 bo for the inside region?
For the region outside the pipe, note that s can become arbitrarily large. What does that tell you about bo for the outside region?

To find ao, 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 an, multiply both sides of the equation by cos(nφ) before integrating.
 

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