A parallel plate capacitor retains its charge (Q) even after being disconnected from a battery, resulting in no immediate loss of charge. The voltage (V) across the capacitor is determined by the relationship Q = C*V, where C is the capacitance. When a dielectric is inserted, the capacitance changes, which affects the voltage, but the charge remains constant unless there is a discharge path. This understanding is crucial for calculating work done during dielectric insertion using the formula E=Q^2/2C.
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xchococatx said:If there is a parallel plate capacitor, connected to a battery, which is then disconnected, there is NO voltage now, but "Q" (charge) on the capacitor remains the same. YES OR NO? I think yes.
xchococatx said:Well basically I'm thinking that a capacitor stores charges (or energy) so, onces its all charged up by a battery, I don't see how it could lose those charges...even when the battery is disconnected. (immediately anyways, it wouldn't be all gone I don't think) Why I was thinking about this was because if the battery is now disconnected, and you wanted to calculate the work needed to insert a dielectric, fitting perfectly between the 2 plates, since there is no voltage, I don't see how you could calculate the work unless you used the original charge of the capacitor (without dielectric). (By means of E=Q^2/2C) So in order to use the original charge of the capacitor, I figured charge has to stay in the capacitor even if the battery is disconnected.