A capacitor has a capacitance of 4.50 μF. What charge would be stored if the potential difference is 6V?
By definition of a capacitance,
C=qVC= \frac{q}{V}C=Vq
q=C×V=4.5×10−6×6=27∗10−6Cq = C \times V = 4.5 \times 10^{-6} \times 6 = 27 * 10^{-6} Cq=C×V=4.5×10−6×6=27∗10−6C
q=27μCq= 27 \mu Cq=27μC
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