High School

Two capacitors of [tex]6.00 \, \text{F}[/tex] and [tex]8.00 \, \text{F}[/tex] are connected in parallel. The combination is then connected in series with a [tex]12.0 \, \text{V}[/tex] battery and a [tex]14.0 \, \text{F}[/tex] capacitor. What is the charge on the [tex]6.00 \, \text{F}[/tex] capacitor?

Answer :

The charge on the 6.00 F capacitor is 72.0 C.

What is the charge on the 6.00 F capacitor?

When two capacitors are connected in parallel, their equivalent capacitance is given by the sum of their individual capacitances. In this case, the equivalent capacitance is 6.00 F + 8.00 F = 14.00 F.

When the parallel combination is connected in series with a 12.0-V battery and a 14.0-F capacitor, the total charge on the parallel combination and the 14.0-F capacitor must be the same. This is because there is no way for charge to escape from the circuit. Thus, we can use the equation Q = CV to find the charge on the 14.0-F capacitor.

Q = CV = (14.0 F)(12.0 V) = 168 C

Now, we can use the fact that the charge on the parallel combination of capacitors is the same to find the charge on the 6.00-F capacitor. Since the equivalent capacitance of the parallel combination is 14.00 F and the voltage across it is 12.0 V, we have:

Q = CV = (14.0 F)(12.0 V) = 168 C

The charge on the 6.00-F capacitor is therefore:

Q = CV = (6.00 F)(12.0 V) = 72.0 C

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