Change in electric potential energy

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SUMMARY

The discussion centers on calculating the change in electric potential energy when a conducting tire's radius changes from 32.5 cm to 10.5 cm, with a charge of 4.89×10-8 C placed on its surface. The initial potential is 1.35E+03 V, and the final potential is 4.19E+03 V. The formula used is ΔU = q(ΔV), leading to a calculated change of 1.4e-4 J. However, the user reports that this calculation does not yield the expected result, indicating a potential error in the application of the formula or values used.

PREREQUISITES
  • Understanding of electric potential and potential energy
  • Familiarity with the concept of charge and its units (Coulombs)
  • Knowledge of the formula for change in electric potential energy
  • Basic grasp of conducting surfaces and their properties
NEXT STEPS
  • Review the principles of electric potential and energy in electrostatics
  • Study the relationship between charge, potential, and energy using examples
  • Examine the effects of changing radius on electric potential in conductors
  • Explore common pitfalls in calculations involving electric potential energy
USEFUL FOR

Students in physics, particularly those studying electrostatics, as well as educators looking to clarify concepts related to electric potential energy and charge interactions.

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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 balloon'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 balloon's surface is 4.19E+03 V.

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

Homework Equations



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

The Attempt at a Solution



\DeltaU = q(\DeltaV) = (4.89e-8C)(4190J/C-1350J/C) = 1.4e-4 J

But that doesn't work.
 
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