91影视

a) We are given four capacitors, where the potential difference across ab is $V_{ab}=50.0V$. The capacitors have a capacitance and localid="1664254520213" $C_1=10渭F,C_2=5渭F,C_3=8渭F{C}_4=9渭F$.

We combine these capacitors to find the net equivalent capacitance In the Figure we see two capacitors C and Cs are in parallel, and we can use the equation of series connection to replace them with their equivalent capacitancelocalid="1664254524423" ${C^{\text{'}}}_{eq}$as next andlocalid="1664254527822" ${C^{\text{'}}}_{eq}$will be as shown in the figure below

As shown in the figure below, we have three capacitors in series, so we can use the equation of the combination of capacitors to replace them with the equivalent capacitance for the system as next

Hence, the equivalent capacitance between and b is 3.47 .$渭F$

b) We are given the potential difference$V_{ab}$ across the entire network, so we can use the value of Ce to get the charge stored in the system by using the equation in the next form

$Q_{net}=C_{eq}{V}_{ab}$ -> (2)

Now we can plug our values forandinto equation (2) to get met

Hence, the charge stored in the capacitor is 174.0 $渭C$.

c) We want to find the charges in capacitors $C_1$ and $C_4$. If we back we will find that the total charge on the lower plate of C and the upper plate of C, together must always be zero because these plates are not connected to anything except each other. Thus in a series connection, the magnitude of the charge on all plates is the same.

So, in our problem, the capacitors$C_1$ and$C_4$ are connected in series with the equivalent capacitor$C{\text{'}}_{eq}$ that we get in part (a). Therefore, the charges on both capacitors are the same for the entire system and equals

$Q=174.0渭C$

Hence, the charge stored in between$C_1$ androle="math" localid="1664253968324" $C_4$ is 174.0 $渭C$.