Change in electric potential energy

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The discussion revolves around calculating the change in electric potential energy when a charged conducting tire shrinks in radius. The initial potential on the tire's surface is 1.35 kV, and after shrinking to a radius of 10.5 cm, the potential increases to 4.19 kV. The formula used for the change in electric potential energy is ΔU = q(ΔV), where q is the charge and ΔV is the change in potential. The user calculated ΔU as 1.4e-4 J but found the result unsatisfactory. Assistance is requested to resolve the discrepancy in the calculation.
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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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