When a battery is connected to the plates of a 3.00-µF capacitor, it stores a charge of 27.0 µC. What is the voltage of the battery?
Q=27"\\mu C" = "27 \\times 10^{-6}" C
C= -3.0 "\\mu F" = -3.0 "\\times 10^{-6}" F
V= voltage of the battery, C= system capacitance , Q= charge stored by the capacitor
"Q= CV"
"V=\\frac{Q}{C}= \\frac{27\\times 10^{-6} C}{3.0\\times 10^{-6}F}"
V=9.0V
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