we have learned that the change in the potential energy of a capacitor when it changes from a neutral condition to one where the plates are charged to Q is predicted by the following equations: DeltaPEQC and DeltaPECDeltaV Where C is the capacitance and DeltaV is the final potential difference between the plates. Explain why, in the first equation, the change in PE is greater the smaller is the capacitance whereas in the second equation, the PE is greater the greater is the capacitance. Although your answer can include purely algebraic reasoning, be sure to include physical reasoning as well.

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In the first equation DeltaPE Q2 2C the change in potential energy of a capacitor is directly proportional to the square of the charge Q and inversely ... View the full answer