Repeat Question 5 for${\mathit{C}}_{\mathbf{2}}$added in series rather than in parallel.

The capacitance $C_1$and $C_2$are in series.

Using the properties of series circuits and using Eq.25-19 and 25-1, we can predict the potential difference across capacitor 1, the charge on capacitor 1, and compare it with its previous value. Also, we can find the equivalent capacitance of capacitor 1 and capacitor 2 using the corresponding formula. Thus, this can determine whether it is more than, less than, or equal to capacitor 1, and the charge stored on capacitor 1 and capacitor 2 together can be found using Eq.25-1 and we can determine whether it is more than, less than, or equal to the charge stored previously on capacitor 1.

Formulae:

If capacitors are in series, the equivalent capacitance$C_{eq}$is given by,

$\frac{1}{C_{eq}}=\sum _{}\frac{1}{C}$ …(i)

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

Considering the properties of the series circuits, we can note that the potential difference is less than previously across $C_1$.

Using equation (ii), we can see that charge is directly proportional to capacitance and voltage. Here,the potential difference$V_1$across$C_1$ is less than its previous value.

Hence, we can say that the $q_1$on$C_1$ is also less than its previous value.

From equation (i), the equivalent capacitance of this series connection is given by:

$\frac{1}{C_{12}}=\frac{1}{C_1}+\frac{1}{C_2}$

Therefore, the equivalent capacitor $C_{12}$of $C_1$and$C_2$ is less than $C_1$.

From calculations of part (c), we can note that iftheequivalent capacitor$C_{12}$of$C_1$and$C_2$ is less than$C_1$ then from equation (ii), we get that the charge is less.

Hence, the charge on $C_1$and $C_2$combined is also less than previously on $C_1$.