91Ó°ÊÓ

We are given a system of three capacitors where ${\mathrm{C}}_1$=6.00$\mathrm{μ}F$and $C_2$=3.00 $\mathrm{μ}F$where both capacitors ${\mathrm{C}}_1\&{\mathrm{C}}_2$are in parallel connection. Also, the third capacitor ${\mathrm{C}}_2$=5.00 $\mathrm{μ}F$where $C_3$is connected in series with the equivalent capacitance $C_{eq}$of $C_1\&C_2$

We are given Q2, so we can use it for Q1 where we see two capacitors ${\mathrm{C}}_1\&{\mathrm{C}}_2$are in parallel, and as the two capacitors are in parallel, hence the two capacitors have the same voltage

$$\begin{gathered} V_1=V_2 \\ \frac{Q_1}{C_1}=\frac{Q_2}{C_2} \\ {Q}_1=Q_2\left(\frac{C_1}{C_2}\right) \end{gathered}$$

Substitute the given values in the above equations.

$Q_1=Q_2\left(\frac{C_1}{C_2}\right)$

$\left(30.0\mathrm{μ}C\right)\left(\frac{6.00\mathrm{μ}E}{3.00\mathrm{μ}E}\right)$

We would get the total charge on the capacitors $C_1\&C_2$because it will help us to find Q3 So the total charge for both capacitors is

$Q=Q_1+Q_2=60.0\mathrm{μ}C+30.0\mathrm{μ}C=90.0\mathrm{μ}C$

As we mentioned in the subsection given the equivalent capacitance of capacitors $C_1\&C_2$is in a series connection with the third capacitor $C_3$wherein a series connection the magnitude of the charge on all plates is the same, therefore, the capacitor $C_3$has the same charge Q of the equivalent capacitance

$Q_3=90.0\mathrm{μ}C$

To determine the applied voltage across ab

$V_{ab}=\frac{Q}{C_{eq}}$

Where Q represents the charge on the third capacitor Q=90.0 $C_{eq}$and is the equivalent capacitance on the system (between $C_{eq}and{C}_{eq}$) in series. In series, the equivalent capacitance $C_{eq}$could be

$\frac{1}{C_{eq}}=\frac{1}{C_{eq}}+\frac{1}{C3}$

Where $C_{eq}$represents the equivalent capacitance between $C_1\&C_2$and both are in parallel

Hence,

$C{\text{'}}_{eq}=C_1+C_2$

Now substitute in the above equation

role="math" localid="1665029468212" $\frac{1}{C_{eq}}=\frac{1}{C_1+C_2}+\frac{1}{C_3}$

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{6.00\mathrm{μ}F+3.00\mathrm{μ}F}+\frac{1}{5.00\mathrm{μ}F} \\ \frac{1}{C_{eq}}=0.311 \\ \frac{1}{C_{eq}}=3.21\mathrm{μ}F \end{gathered}$$

Hence the potential difference across ab is

$V_{ab}=\frac{Q}{C_{eq}}=\frac{90.0\mathrm{μ}C}{3.21\mathrm{μ}F}=28.0V$