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Let's say I turn on a Van de Graaff which creates a large positive charge. Now let's say I have an object with a positive charge in my hand and I start walking toward the Van de Graaff from $x$ meters away. If I walk an arbitrary distance towards the Van de Graaff, I am doing work and the charge in my hand is gaining potential energy. Now somebody standing next to the Van de Graaff suddenly turns it off. What happens to that potential energy of the charge? By the conservation of energy it should be conserved somehow, but it appears not to be. Any help is appreciated.

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The Van de Graaff generator itself has stored potential energy, even without the extra charged object. Like a capacitor, there will be energy stored in the electric field. When you turn the generator off, this energy doesn't vanish by itself but either dissipates slowly through corona discharge, or quickly by arc discharge.

The potential energy in moving a charged object close to the generator would go through the same pathway. By bringing the object close to the generator, you will alter the electric field, increasing the electric potential on the generator. More energy will then be dissipated in any arc or corona discharge.

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Thanks a lot, that completely makes sense! –  Ovi Mar 18 '13 at 4:42

To keep the object in a fixed location (assume you are walking very slowly) while the field was on, you needed a force. When the field is turned off this force (your hand, having potential energy like a loaded spring) will perform work and generate kinetic energy of the object.

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