Figure 25-54 shows capacitor 1 $(C_1=8.00\mathrm{\mu F})$, capacitor 2 $\mathbf{(}{\mathbf{C}}_{\mathbf{2}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu F}\mathbf{)}$, and capacitor 3$\mathbf{(}{\mathbf{C}}_3\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu F}\mathbf{)}$ connected to a 12.0 V battery. When switch S is closed so as to connect uncharged capacitor 4 $\mathbf{(}{\mathbf{C}}_4\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu F}\mathbf{)}$, (a) how much charge passes through point Pfrom the battery and (b) how much charge shows up on capacitor 4? (c) Explain the discrepancy in those two results.

1. Potential difference of a battery $V=12V$,
2. Capacitance of capacitor 1,$C_1=8\mathrm{\mu F}$
3. Capacitance of capacitor 2,$C_2=6\mathrm{\mu F}$
4. Capacitance of capacitor 3,$C_3=8\mathrm{\mu F}$
5. When switch S is closed, the additional capacitance of capacitor 4 is $C_4=6\mathrm{\mu F}$.

For solving this problem, we have to use the formula for the equivalent capacitance for series and parallel combinations. By using the relation between capacitance and charge, we can find the charge that passes through the point and also the charge on the capacitor.

Formulae:

The equivalent capacitance of a series connection of capacitors,

$\frac{1}{C_{\mathrm{equivalent}}}=\sum _{}\frac{1}{C_j}$ …(i).

The equivalent capacitance of a parallel connection of capacitors,

$\frac{1}{C_{\mathrm{equivalent}}}=\sum _{}{C}_j$ …(ii)

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

Initially, the three capacitors i.e. $C_1,C_2$and$C_3$ are in series; so their equivalent capacitance can be calculated by using the given data in equation (i) as follows:

$$\begin{gathered} \frac{1}{C_{123}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \\ \frac{1}{C_{123}}=\frac{C_1{C}_2+C_2{C}_3+C_1{C}_3}{C_1{C}_2{C}_3} \\ \frac{1}{C_{123}}=\frac{C_1{C}_2{C}_3}{C_1{C}_2+C_2{C}_3+C_1{C}_3} \\ =\frac{\left(8\mathrm{\mu F}\times 6\mathrm{\mu F}\times 8\mathrm{\mu F}\right)}{\left(8\mathrm{\mu F}\times 6\mathrm{\mu F}\right)\left(6\mathrm{\mu F}\times 8\mathrm{\mu F}\right)\left(8\mathrm{\mu F}\times 8\mathrm{\mu F}\right)} \\ =\frac{348{\left(\mathrm{\mu F}\right)}^3}{\left(48+48+64\right){\left(\mathrm{\mu F}\right)}^2} \\ =\frac{384{\left(\mathrm{\mu F}\right)}^3}{160{\left(\mathrm{\mu F}\right)}^2} \\ =2.4\mu F \end{gathered}$$.

Now, we can obtain the total initial charge on the equivalent capacitance $C_{123}$by using the given data in equation (iii) as follows:.

$$\begin{gathered} q_i=\left(2.4\mathrm{\mu F}\right)\times \left(12\mathrm{V}\right) \\ =28.8\mathrm{\mu C} \end{gathered}$$ …(iv)

Finally,$C_2$and$C_4$are in parallel combination, so the equivalent capacitance is given by using the given data in equation (ii) as follows:

$$\begin{gathered} C_{24}=6\mu F+6\mu F \\ =12\mu F \end{gathered}$$

So, now the equivalent capacitance of the circuit is given using equation (i) for the series combination as follows:

$$\begin{gathered} \frac{1}{C_{1234}}=\frac{1}{C_1}+\frac{1}{C_{24}}+\frac{1}{C_3} \\ \frac{1}{C_{1234}}=\frac{C_1{C}_{24}+C_{24}{C}_3+C_1{C}_3}{C_1{C}_{24}{C}_3} \\ \frac{1}{C_{1234}}=\frac{C_1{C}_{24}{C}_3}{C_1{C}_{24}+C_{24}{C}_3+C_1{C}_3} \\ =\frac{\left(8\mathrm{\mu F}\times 12\mathrm{\mu F}\times 8\mathrm{\mu F}\right)}{\left(8\mathrm{\mu F}\times 12\mathrm{\mu F}\right)\left(12\mathrm{\mu F}\times 8\mathrm{\mu F}\right)\left(8\mathrm{\mu F}\times 8\mathrm{\mu F}\right)} \\ =\frac{768{\left(\mathrm{\mu F}\right)}^3}{256{\left(\mathrm{\mu F}\right)}^2} \\ =3\mathrm{\mu F} \end{gathered}$$

Therefore, the final charge is given using equation (iii) as:

$$\begin{gathered} q_f=3\mathrm{\mu F}\times 12\mathrm{V} \\ =36\mathrm{\mu F} \end{gathered}$$ …(v)

The charge that passes through the point P is the change in the initial and the final charge. Thus, from equations (iv) and (v), the require amount of charge can be calculated as:

$$\begin{gathered} ∆q=36\mathrm{\mu C}-28.8\mathrm{\mu C} \\ =7.2\mathrm{\mu C} \end{gathered}$$

Hence, the value of the charge is $7.2\mathrm{\mu C}$.

Since the capacitor $C_{24}$is in series with capacitor $C_1$and $C_3$. So, the final charge on the capacitors is same.

Thus, the potential across the $C_{24}$can be given using equation (iii) as follows:

$$\begin{gathered} V_{24}=\frac{36\mathrm{\mu C}}{12\mathrm{\mu F}} \\ =3\mathrm{V} \end{gathered}$$

This voltage is same across each of the parallel capacitors. So, the voltage across the capacitor 4 is $V_4=3V$

Thus, the charge on capacitor 4 is given using equation (iii) as follows:

$$\begin{gathered} q_4=6\mathrm{\mu F}\times 3\mathrm{V} \\ =18\mathrm{\mu C} \end{gathered}$$

Hence, the value of the charge is $18\mathrm{\mu C}$.

The battery supplies the charges only to the plates where it is connected. According to the new distribution of voltages across the capacitors, the charges on rest of the plates are due to electron transfer between them. The battery does not directly supply the charge on capacitor 4.