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How much charge is on each plate of a 4.00-F capacitorwhen the is associated to a 12.0-V battery?

I said 2.4 x 10^-5 C since there are two key of a parallel plate capacitor. But the vital said just 4.8 x 10^-5 C




The fee on the two plates is opposite, and also equal to the an essential answer. The parameter Q is the charge on one of the plates, not the amount of the absolute worth of the two charges. The meaning of capacitance is the fee per unit voltage, so that $Q=CV$, for this reason 4F times 12V is 48 systems of charge, which for the unit "F" is millionths the a Coulomb C. This provides the book"s answer.


The formula $V=fracQC$ offers the lot of charge that is there on one of the plates. The total amount of charge on both the plates taken together is zero! Both the them space oppositely charged.



Net charge on capacitor is constantly zero since there is equal and also unlike dues on plates.Hence capacitor is not charge save device. The is electrical energy save on computer device.

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In any type of capacitor, stored charge when charged through voltage V is q=cvwhere +cv is save on computer in one plate and -cv is save on computer in one more plate.

In this inquiry charge stored have to be: q = 4*10^-6 * 12 coulombi.e q = 48 micro coulomb

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