Electric Potential in Insulated U-Tube

In summary, the potential at the center of curvature of the U-tube can be calculated by breaking the problem into two parts (arc and arms), using the equation V = (1/4πε₀) (Q/r) for each part, and then summing up the resulting potentials.
  • #1
PK_Kermit
1
0
Hey, I have this question, and I'm having quite a bit of difficulty figuring it out.

"An insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along the rod. Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity."

Which i see to be as so:

...|...|..........
...|...|...........
...|...o...|...<- the 'o' represents the center of curvature of
...\... /.........the arc...
....\.../...<-- this part is round.....
...u............

There is a similar question in my textbook, where tehy work out the potential at a point from a ring of charge. they use the equasion:

V= (1/4[pi]E0) [integral] dq/r

[where dq would be infinitismal elements of charge about the ring]

for the ring, the r distance is always the same; so they remove r and integrate dq, to end with:

V= (1/4[pi]E0) ( Q / r )

now, my question is the U tube. I figure, since under the center of curvature lies a semicircle, maybe you could take half the potential found from the previous equasion; i.e., for the curvature part, you could use:

0.5(1/4[pi]E0)( Q / r )

but, there is nothing in my textbook to say what I would do in terms of the extending arms of the U tube; as well, the question doesn't say anything about the length of the arms.

what is the significance of defining V to be zero at infinity, and where should I go from here?

Thanks,
PK_Kermit
 
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  • #2
The significance of defining the potential to be zero at infinity is that it allows you to calculate the potential at any given point using the principle of superposition. This means that you can break down the problem into parts, and sum up the resulting potentials at each part to get the total potential at the center of curvature. For this particular problem, you can calculate the potential at the center of curvature by breaking the problem into two parts: the potential due to the arc of the U-tube and the potential due to the two arms of the U-tube. The potential due to the arc can be calculated using the equation you provided (V = (1/4πε₀) (Q/r), with r being the radius of the arc and Q being the total charge). For the arms, you can use the same equation, but with the distance from the center of curvature to the end of the arm as the value for r. Once you've calculated the potential due to each part, you can add them together to get the total potential at the center of curvature.
 
  • #3


Hi PK_Kermit,

First, let's address the significance of defining the potential to be zero at infinity. This is a common convention used in electrostatics to make calculations easier. Since the potential at infinity is assumed to be zero, it means that any point at a finite distance from the charge will have a potential that is greater than zero. This allows us to focus on the relative differences in potential between different points, rather than having to deal with absolute values.

Now, let's move on to the problem at hand. You are correct in thinking that you can use the equation V= (1/4[pi]E0)( Q / r ) for the semicircular arc of the U-tube. However, you cannot simply take half of this value for the entire U-tube. This is because the potential at the center of curvature is not the same as the potential at points along the extended arms of the U-tube.

To find the potential at the center of curvature, you can use the equation you mentioned and integrate over the entire length of the semicircular arc. This will give you the potential at the center of curvature due to the charge on the semicircular arc. To find the potential at points along the extended arms, you can use the same equation and integrate over the length of the arms, using the distance from the charge to each point as the value for r.

Once you have calculated the potential at different points along the U-tube, you can add them together to find the total potential at any given point. Keep in mind that the potential at any point along the extended arms will also be affected by the potential at the center of curvature, so you will need to take this into account when adding the potentials together.

I hope this helps guide you in the right direction. Remember to always consider the significance of the given information and use the correct equations to solve the problem. Good luck!
 

Question 1: What is an insulated U-tube?

An insulated U-tube is a scientific apparatus used to demonstrate the principles of electric potential. It consists of two parallel glass tubes connected at the top and bottom, and filled with an insulating liquid such as oil or water. The tubes are also connected to a power source, allowing for the creation of an electric potential difference between the two tubes.

Question 2: How does an insulated U-tube demonstrate electric potential?

The insulated U-tube demonstrates electric potential by showing the movement of charged particles within the liquid. When a potential difference is created between the two tubes, one tube becomes positively charged and the other becomes negatively charged. This causes the charged particles within the liquid to move, creating a visible difference in the liquid levels between the two tubes.

Question 3: What factors affect the electric potential in an insulated U-tube?

The electric potential in an insulated U-tube is affected by the distance between the two tubes, the type of insulating liquid used, and the amount of charge applied to the tubes. Additionally, the shape and material of the tubes can also play a role in the electric potential observed.

Question 4: What is the purpose of using an insulating liquid in an insulated U-tube?

The insulating liquid serves to prevent the flow of electricity between the two tubes, allowing for a clear demonstration of electric potential. If a conductive liquid were used, the charged particles would simply flow between the tubes, negating the effects of the potential difference.

Question 5: What are some real-world applications of understanding electric potential in an insulated U-tube?

Understanding electric potential in an insulated U-tube can have practical applications in the development of electrical circuits and systems. It can also be used to study the behavior of charged particles and can aid in the development of technologies such as batteries and capacitors. Additionally, the principles demonstrated by an insulated U-tube can also be applied to understanding the behavior of electricity in nature, such as lightning and electrical storms.

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