# Question #78627

Jan 18, 2017

(A)

#### Explanation:

Two given capacitors are connected as shown in figure above.
The total capacitance ${C}_{\text{T}}$ can be found by

${C}_{\text{T}} = {C}_{1} + {C}_{2}$ .........(1)

For a charge $q$ given to the assembly

${C}_{\text{T}} = \frac{q}{V}$
where $V$ is voltage drop across capacitors and is equal for both capacitors.

If ${q}_{1} \mathmr{and} {q}_{2}$ is the charge distributed on capacitors ${C}_{1} \mathmr{and} {C}_{2}$ respectively.

${C}_{1} = {q}_{1} / V$ ......(2) and
${C}_{2} = {q}_{2} / V$ .....(3)

Dividing equation (2) by (3) we get
${C}_{1} / {C}_{2} = \frac{{q}_{1} / V}{{q}_{2} / V}$
$\implies \frac{{q}_{1}}{{q}_{2}} = {C}_{1} / {C}_{2}$