What is the force on a 1000 kg elevator that is free falling at 9.8 m/s/s?
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The electrostatic potential energy, U, of one point charge q at position d in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force to bring it from the reference position d to that position
Thus, to double the electric potential energy U we need to reduce the distance of separation by half (1/2) because they are inversely proportion