Change in electric potential energy

[Physics197]

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

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

But that doesn't work.

 

[Physics197]

anyone have a clue, or a hint