Question #153231

A capacitor has a capacitance of 4.50 μF. What charge would be stored if the potential difference is 6V?



1
Expert's answer
2021-01-04T14:35:31-0500

By definition of a capacitance,

C=qVC= \frac{q}{V}

q=C×V=4.5×106×6=27106Cq = C \times V = 4.5 \times 10^{-6} \times 6 = 27 * 10^{-6} C

q=27μCq= 27 \mu C


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