Figure 25-58 shows a four capacitor arrangement that is connected to a larger circuit at points Aand B. The capacitances are${\mathbf{C}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{\mu F}$and${\mathit{C}}_{\mathbf{2}}\mathbf{=}{\mathit{C}}_{\mathbf{3}}\mathbf{=}{\mathit{C}}_{\mathbf{4}}\mathbf{=}\mathbf{20}\mathbf{}\mathit{\mu }\mathit{F}$. The charge on capacitor 1 is$\mathbf{30}\mathbf{}\mathbf{\mu C}$. What is the magnitude of the potential difference${\mathit{V}}_{\mathbf{A}}\mathbf{-}{\mathit{V}}_{\mathbf{B}}$?

a) Capacitance$C_1=10\mu F$

b) Capacitance$C_2=C_3=C_4=20\mu F$

c) Charge on capacitor 1$q_1=30\mu C$

We first find the voltage across capacitor 1 using the charge and the capacitance on it. Since the combination of capacitors 1 and 2 is parallel, the potential of both the capacitors is the same. Using this idea, we find the charge of capacitor 2. Now, the total charge of capacitors 1 and 2 is also the charge of the branch formed by capacitors 3 and 4. Since the capacitance of 3 and 4 is the same, the charge on each of these capacitors is also the same. Using this, we find the potential difference between them. So, finally, we can find the required result.

Formula:

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

The voltage across the capacitor 1 can be calculated using equation (i) as:

$$\begin{gathered} V_1=\frac{q_1}{C_1} \\ =\frac{30\mathrm{\mu C}}{10\mathrm{\mu F}} \\ =3.0V \end{gathered}$$

Since the capacitors 1 and 2 are parallel, the potential difference across the capacitors 1 and 2 is the same. Thus,. The value of potential difference across the capacitor 2 is given as:

$$\begin{gathered} V_1=V_2 \\ =3.0V \end{gathered}$$

The total charge on the capacitor 2 now, is given using equation (i) as:

$$\begin{gathered} q_2=\left(20\mu F\right)\left(3V\right) \\ =60\mu C \end{gathered}$$

Which means a total of$90\mu C$ of charge is on the pair of capacitors$C_1$and $C_2$. This means that the same charge will be on the pair of$C_3$and$C_4$Also $C_3=C_4$, then the charges equally divide between them.

$$\begin{gathered} q_3=q_4 \\ =\frac{90\mu C}{2} \\ =45\mu C \end{gathered}$$

And the voltage across capacitor 3 is now given using equation (i) as:

$$\begin{gathered} V_3=\frac{q_3}{C_3} \\ =\frac{45\mu C}{20\mu F} \\ =2.3V \end{gathered}$$

Therefore, the potential difference across the point A and Bpointis given as:

$$\begin{gathered} V_A-V_B=V_1+V_3 \\ =3.0V+2.3V \\ =5.3V \end{gathered}$$

Hence, the required potential difference value is 5.3 V.