Figure 25-20 shows three circuits, each consisting of a switch and two capacitors, initially charged as indicated (top plate positive). After the switches have been closed, in which circuit (if any) will the charge on the left-hand capacitor (a) increase, (b) decrease, and (c) remain the same?

Fig.25-20 with three circuits, each of a switch and two capacitors is given.

By using Eq.25-1, we can find the potential difference across each capacitor. The charge moves from higher potential to lower potential. Using this, we can find the circuit in which the charge on the left-hand capacitor increases, decreases, and remain the same.

Formula:

The charge within the plates of the capacitor, $q=CV$ …(i)

Thus, the potential difference using equation (i) is given as follows:

$V=\frac{q}{C}$ …(ii)

Putting the given values in equation (ii), the potential difference on the left hand capacitor for the circuit 1 is given as:

$$\begin{gathered} V_{L1}=\frac{6q}{2C} \\ =\frac{3q}{C} \end{gathered}$$

On the right hand capacitor, the potential on circuit 1 is given as:

$V_{R1}=\frac{3q}{C}$

Putting the given values in equation (ii), the potential difference on the left hand capacitor for the circuit 2 is given as:

$V_{L2}=\frac{6q}{3C}=\frac{2q}{C}$

On the right hand capacitor, the potential on circuit 2 is given as:

$V_{R2}=\frac{3q}{C}$

Putting the given values in equation (ii), the potential difference on the left hand capacitor for the circuit 3 is given as:

$$\begin{gathered} V_{L3}=\frac{6q}{2C} \\ =\frac{3q}{C} \end{gathered}$$

On the right hand capacitor, the potential on circuit 3 is given as:

$V_{R3}=\frac{3q}{2C}$

We know that the chargemoves from higher potential to lower potential.

Thus,from the calculations of part (a)in circuit 2, we can see that $V_{L2}<V_{R2}$.

Hence, here the charge on the left-hand capacitor increases.

Similarly from the calculations of part (a)in circuit 3, we can see that $V_{L3}<V_{R2}$.

Hence, here the charge on the left-hand capacitor decreases.

Similarly, from the calculations of part (a) in circuit 1, we can see that $V_{L1}<V_{R1}$.

Hence, here is no change in the charge on the left-hand capacitor.