91Ó°ÊÓ

1. Capacitance,${\mathrm{C}}_1=4.00\mathrm{μ¹ó}$
2. Capacitance,${\mathrm{C}}_2=6.00\mathrm{μ¹ó}$
3. Capacitance,${\mathrm{C}}_3=3.00\mathrm{μ¹ó}$
4. Voltage,$\mathrm{V}=12.0\mathrm{V}$

We find the equivalent capacitance of capacitors 2 and 3 first, and then, using the condition that the potential difference across capacitors 1, 2, and 3 is the same and using conservation of charge, we calculate the charge on capacitor 1 and finally for 2 and 3.

Formulae:

The equivalent capacitance of a series connection of capacitors,

$\frac{1}{{\mathrm{C}}_{\mathrm{equivalent}}}=\underset{}{∑}\frac{1}{{\mathrm{C}}_{\mathrm{i}}}$ ...(i).

The equivalent capacitance of a parallel connection of capacitors,

${\mathrm{C}}_{\mathrm{equivalent}}=\underset{}{∑}{\mathrm{C}}_{\mathrm{i}}$ ...(ii)

The charge stored between the plates of the capacitor, q = CV ...(iii)

The charges on capacitor 2 and 3 are same because they are connected in series and replaced by equivalent capacitance. Thus, the equivalent capacitance is given using equation (i) as:

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{C_2}+\frac{1}{C_3} \\ =\frac{C_2+C_3}{C_2{C}_3} \end{gathered}$$ …(¾±±¹)

And the potential difference across the equivalent capacitor is given using equation (iii) as follows:

$V_2=\frac{q_2}{C_{eq}}$

The potential difference on capacitor 1 is given using equation (iii) as follows:

$V_1=\frac{q_1}{C_1}$

This equivalent capacitor and capacitor 1 are now in parallel combination. The potential difference across both the capacitances is same. Thus, the equation of charge and capacitance is given as:

$\frac{q_1}{C_1}=\frac{q_2}{C_{eq}}$ …(±¹)

Now suppose the original chargeon capacitor 1 flows to the combination of capacitor 2 and 3.

Using conservation of charge and equation (iii), we get that

$$\begin{gathered} q_1+q_2=q_0 \\ =C_1{V}_0 \end{gathered}$$

…(±¹¾±)

where,${\mathrm{V}}_0$is the original potential difference across capacitor 1.

Solving both the equations (v) and (vi), we can get the charge as follows:

$q_2=\frac{q_1}{C_1}{C}_{eq}$

Substituting this value in equation (vi), we get that

$$\begin{gathered} q_1=\frac{q_1}{C_1}{C}_1{V}_0 \\ {q}_1\left(1+\frac{C_{eq}}{C_1}\right)=C_1{V}_0 \\ \frac{q_1\left(C_1+C_{eq}\right)}{C_1}=C_1{V}_0 \\ {q}_1=\frac{C_1^2{V}_0}{\left(C_1+C_{eq}\right)} \end{gathered}$$

Substituting value of equation (iv) in the above value, we get the charge value on capacitor 1 as:

$$\begin{gathered} {\mathrm{q}}_1=\frac{{\mathrm{C}}_1^2{\mathrm{V}}_0}{{\displaystyle \frac{{\mathrm{C}}_2{\mathrm{C}}_3}{{\mathrm{C}}_2+{\mathrm{C}}_3}}+{\mathrm{C}}_1} \\ =\frac{{\mathrm{C}}_1^2\left({\mathrm{C}}_2+{\mathrm{C}}_3\right){\mathrm{V}}_0}{{\mathrm{C}}_1{\mathrm{C}}_2={\mathrm{C}}_1{\mathrm{C}}_3+{\mathrm{C}}_2{\mathrm{C}}_3} \\ =\frac{{\left(4.00\mathrm{μ¹ó}\right)}^2\left(6.00\mathrm{μ¹ó}+3.00\mathrm{μ¹ó}\right)\left(12.0\mathrm{V}\right)}{\left(24+12+18\right){\left(\mathrm{μ¹ó}\right)}^2} \\ =32.0\mathrm{μ°ä} \end{gathered}$$.

Hence, the value of the charge is$32.0\mathrm{μ°ä}$ .

Charge on the capacitor 2 is given by substituting the charge of capacitor 1 in equation (c) as follows:

$$\begin{gathered} {\mathrm{q}}_2={\mathrm{C}}_1{\mathrm{V}}_0-{\mathrm{q}}_1 \\ =\left(4.00\mathrm{μ¹ó}\right)\left(12.0\mathrm{V}\right)-32.0\mathrm{μ°ä} \\ =16.0\mathrm{μ°ä} \end{gathered}$$

Hence, the value of the charge is $16.0\mathrm{μ°ä}$.

The charge on the capacitor 3 is the same as capacitor 2 because they are connected in a series.

Hence, the value of the charge is $16.0\mathrm{μ°ä}$.