91Ó°ÊÓ

The capacitance $C_1$and $C_1$are in parallel.

Using the properties of parallel circuits and using Eq.25-19 and 25-1, we can predict the potential difference across capacitor 1, the charge 1 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:

The charge within the plates of the capacitor,$q=CV$ …(¾±)

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

$C_{eq}=∑C$ …(¾±¾±)

Considering the properties of the parallel circuits, the potential remains unchanged in the parallel capacitors.

Therefore, we can note that the potential difference is the same than previously across $C_1$.

The properties of parallel circuits give the charges that remain the same in the circuit considering equation (i).

Therefore, the charge $q_1$on $C_1$is now the same as previous.

From equation (ii), the equivalent capacitance of this parallel connection is given by:

$C_{12}=C_1+C_2$

Therefore, capacitance is increased after parallel arrangement. This implies that, the equivalent capacitor $C_{12}$of $C_1$and$C_2$ is more than $C_1$.

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

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