Potential inside a rectangular pipe

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The discussion centers on the independence of electric potential within a rectangular pipe aligned along the z-axis, emphasizing that this potential does not vary with the z-coordinate. The uniqueness theorem is mentioned as a possible explanation for this phenomenon. Participants highlight the role of symmetry, noting that every point along the axis is equivalent, leading to uniform potential. The conversation suggests that understanding these principles is crucial for grasping the behavior of potential in such geometries. Overall, the independence of potential is attributed to both symmetry and theoretical foundations in electrostatics.
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TL;DR Summary: Independence of potential( inside a rectangular pipe running along z axis)from z coordinate

Consider the following diagram

Screenshot_2023-06-08-15-34-01-36_e2d5b3f32b79de1d45acd1fad96fbb0f.jpg

It is an infinite rectangular pipe running along z axis.I know that the potential inside the pipe is independent of z coordinate, but I cannot seem to convince myself of it.My guess is that it has to do something with uniqueness theorem.
 
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Welcome to PF. I've moved your thread to the schoolwork forums.

What equations would apply to this problem?
 
better said:
My guess is that it has to do something with uniqueness theorem.
Symmetry, I'd have said. Everywhere along the axis looks like everywhere else.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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