Potential difference of a conducting pipe

In summary, the potential difference between the axis and outer surface of a conducting cylindrical pipe with a charge q released inside is infinite. This is due to the fact that the electric field is strongest at the surface of a conductor and the potential difference is directly proportional to the electric field. The integral should be bounded between 0 and L, and the natural logarithm function should be used when integrating 1/r.
  • #1
Suziii
4
0

Homework Statement



A conducting cylindrical pipe has an inner radius Ra and an outer radius Rb. a charge q is released inside it. Assume the length of a cylinder L>>Rb. Find the potential difference between the axis of the cylinder and outer surface.

Homework Equations



Vb-Va=-∫E*dl (vector multiplication and the integral is bounded between a and b)

The Attempt at a Solution


I found the electric field in the outer surface to be E=q/ε2πRL
Vb-Va= q/ε2πL ∫1/rdr= q/ε2πL (lnRb-lnRa) (note: the integral is definite between Ra and Rb)
Since the question asks for the potential difference between the axis and the outer surface I need to run R between 0 and R, so i get a ln(0) which does not make sense...

Pleaseeeeeeeeee helppppp
 
Physics news on Phys.org
  • #2

Thank you for your question. Your solution is on the right track, but there are a few things that need to be clarified.

First, you are correct in finding the electric field on the outer surface to be E=q/ε2πRL. However, this is only true for points on the outer surface, not points on the axis. For points on the axis, the electric field will be zero since the charge is located at the center of the cylinder and there is no electric field inside a conductor.

Next, your integral should not be bounded between Ra and Rb, but rather between 0 and L. This is because the potential difference between the axis and outer surface is the same as the potential difference between any two points on the axis and outer surface, regardless of the distance between them. So, we can choose any two points on the axis and outer surface to find the potential difference, and it will be the same as the potential difference between the axis and outer surface.

Finally, when integrating, you should use the natural logarithm function (ln), not log base 10. Also, when integrating 1/r, you should use the limit as r approaches 0, not the limit as r approaches R.

So, the correct solution would be:

Vb-Va= q/ε2πL ∫1/rdr= q/ε2πL (lnL-ln0) = q/ε2πL (lnL-(-∞)) = q/ε2πL (lnL+∞) = ∞

This means that the potential difference between the axis and outer surface is infinite. This makes sense because the electric field is strongest at the surface of a conductor, and since the potential difference is directly proportional to the electric field, the potential difference will also be infinite.

I hope this helps clarify your solution. Let me know if you have any further questions.
 

FAQ: Potential difference of a conducting pipe

1. What is potential difference?

Potential difference is the difference in electric potential between two points in a circuit. It is measured in volts (V) and represents the amount of energy required to move a unit of charge from one point to another.

2. What is a conducting pipe?

A conducting pipe is a type of pipe that is made from a material that allows electricity to flow through it easily. Examples of conducting materials include copper, aluminum, and silver.

3. How does potential difference affect a conducting pipe?

Potential difference causes a flow of electrons through a conducting pipe, creating an electric current. The higher the potential difference, the greater the flow of electrons and the stronger the electric current.

4. What factors can affect the potential difference of a conducting pipe?

The potential difference of a conducting pipe can be affected by the material it is made of, the length and thickness of the pipe, and the temperature of the surrounding environment.

5. Why is potential difference important?

Potential difference is important because it is necessary for the flow of electricity in a circuit. It is also a key factor in determining the strength of an electric current and the amount of energy that can be transferred through a conducting pipe.

Back
Top