Question

Question #78627

Answer 1

(A)

Explanation:

Two given capacitors are connected as shown in figure above.
The total capacitance `C_"T"` can be found by

`C_"T"=C_1+C_2` .........(1)

For a charge `q` given to the assembly

`C_"T"=q/V`
where `V` is voltage drop across capacitors and is equal for both capacitors.

If `q_1 and q_2` is the charge distributed on capacitors `C_1 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=(q_1/V)/(q_2/V)`
`=>(q_1)/(q_2)=C_1/C_2`