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How much charge is on each plate of a 4.00-F capacitor when it is connected to a 12.0-V battery?

I said 2.4 x 10^-5 C because there are two plates of a parallel plate capacitor. But the key said only 4.8 x 10^-5 C

Why?

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BTW--Having repeated problems in a single closely defined problem domain suggests that you may be working from a inadequate mental model. It might be worth looking at the basics again or talking to your instructor or trying a different source. –  dmckee May 3 '12 at 16:43
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I don't have an instructor. –  sidht May 3 '12 at 16:53
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@jak get a textbook (I suggest Resnick) and revise this topic along with its foundation topics. Also check out online sources like Khan Academy or MIT OCW. –  Manishearth May 3 '12 at 17:15

3 Answers 3

up vote 2 down vote accepted

The charge on the two plates is opposite, and equal to the key answer. The parameter Q is the charge on one of the plates, not the sum of the absolute value of the two charges. The definition of capacitance is the charge per unit voltage, so that $Q=CV$, so 4F times 12V is 48 units of charge, which for the unit "F" is millionths of a Coulomb C. This gives the book's answer.

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Thank you for your explanation. –  sidht May 3 '12 at 16:54

Net charge on capacitor is always zero because there is equal and unlike charges on plates. Hence capacitor is not charge storing device. It is electrical energy storing device.

In any form of capacitor, stored charge when charged by voltage V is q=cv where +cv is stored in one plate and -cv is stored in another plate.

In this question charge stored should be: q = 4*10^-6 * 12 coulomb i.e q = 48 micro coulomb

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The formula $V=\frac{Q}{C}$ gives the amount of charge that is there on one of the plates. The total amount of charge on both the plates taken together is zero! Both of them are oppositely charged.

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