Change in electric potential energy

  • Thread starter Physics197
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  • #1
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1. Homework Statement

A tire of radius 32.5 cm has a surface that is conducting. A charge of 4.89×10-8 C is then placed on the surface. The potential on the tire's surface is 1.35E+03 V. Later, suppose that some of the air is let out of the tire, so that the radius shrinks to 10.5 cm, the new potential on the tire's surface is 4.19E+03 V.

What is the change in electrical potential energy during the shrinking process?

2. Homework Equations

Change in electric potential energy = q (change in electric potential)

3. The Attempt at a Solution

Delta U = qDelta V) = (4.89e-8C)(4190J/C-1350J/C) = 1.4e-4 J

But that doesnt work.
 

Answers and Replies

  • #2
656
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1. Homework Statement

A tire of radius 32.5 cm has a surface that is conducting. A charge of 4.89×10-8 C is then placed on the surface. The potential on the tire's surface is 1.35E+03 V. Later, suppose that some of the air is let out of the tire, so that the radius shrinks to 10.5 cm, the new potential on the tire's surface is 4.19E+03 V.

What is the change in electrical potential energy during the shrinking process?

2. Homework Equations

Change in electric potential energy = q (change in electric potential)

3. The Attempt at a Solution

Delta U = qDelta V) = (4.89e-8C)(4190J/C-1350J/C) = 1.4e-4 J

But that doesnt work.

There is an equation for V, just electric potential, that involves charge and distance (r).
 
  • #3
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V = KQ/r and then I can solve for the potential for both cases again...

Still doesnt help find the change in potential ENERGY
 
  • #4
656
2
V = KQ/r and then I can solve for the potential for both cases again...

Still doesnt help find the change in potential ENERGY

Ok maybe I am reading this wrong but it says that the surface of the tire already has an electric potential. Then R is decreased, I guess the pressure released does work, and brings the surface to a new potential. So you need to find what the initial charge on the tire was and you can do this because you have V and R.

You know what... I am now not sure.
Its easy if all we are worried about is the point charge on the surface. But I am assuming, maybe incorrectly that the tire already has charge and we are also bringing all those charges in a circle closer together...

Sorry, and I will await anothers response
 

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