(a) When a battery is connected to the plates of a 8.00-µF capacitor, it stores a charge of 48.0 µC. What is the voltage of the battery? V (b) If the same capacitor is connected to another battery and 192.0 µC of charge is stored on the capacitor, what is the voltage of the battery?
Question
Answer:
a.6 V
b.24 V
Explanation:
We are given that
a.[tex]C=8\mu F=8\times 10^{-6} F[/tex]
[tex]1\mu =10^{-6} [/tex]
Q=[tex]48\mu C=48\times 10^{-6} C[/tex]
We know that
[tex]V=\frac{Q}{C}[/tex]
Using the formula
[tex]V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V[/tex]
b.[tex]Q=192\mu C=192\times 10^{-6} C[/tex]
[tex]V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V[/tex]