Question

In Fig., C1 = 6.00 mF, C2 = 3.00 mF, and C3 = 5.00 mF. The capacitor network is connected to an applied potential Vab. After the charges on the capacitors have reached their final values, the charge on C2 is 30.0 mC.

(a) What are the charges on capacitors C1 and C3?

(b) What is the applied voltage Vab?

Short answer

a) The charge on the capacitor ${\mathrm{C}}_1$is 60$\mathrm{\mu F}$.

The charge on the capacitor role="math" localid="1665027698801" $C_3$is 90$\mathrm{\mu F}$

b) The potential difference across a and b is 28.0V.

Step-by-step solution

About connections of capacitors

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

The charge on the second capacitance is $Q_2$=30.0data-custom-editor="chemistry" $\mathrm{\mu F}$.

Charge on each resistors.

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.

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

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

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

$Q=Q_1+Q_2=60.0\mu C+30.0\mu C=90.0\mu C$

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

$Q_3=90.0\mu C$

About potential difference.

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" $\frac{1}{C_{eq}}=\frac{1}{C_1+C_2}+\frac{1}{C_3}$

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

Hence the potential difference across ab is

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