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A capacitor has a capacitance of 5.48 pF. What is the charge on the positive plate...

A capacitor has a capacitance of 5.48 pF. What is the charge on the positive plate when the magnitude of the potential difference between the plates is 13.2 V?

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Answer #1

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Answer #2

To find the charge on the positive plate of the capacitor, we can use the formula for the capacitance of a capacitor:

Q = C * V

Where: Q = Charge on the capacitor (in coulombs) C = Capacitance of the capacitor (in farads) V = Potential difference between the plates (in volts)

Given: Capacitance (C) = 5.48 pF = 5.48 x 10^-12 F (convert from pF to farads) Potential difference (V) = 13.2 V

Now, plug the values into the formula:

Q = 5.48 x 10^-12 F * 13.2 V

Q ≈ 7.2336 x 10^-11 coulombs

The charge on the positive plate of the capacitor is approximately 7.2336 x 10^-11 coulombs.

answered by: Hydra Master
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