Question #78627

1 Answer
A08
Jan 18, 2017

(A)

Explanation:

![http://farside.ph.utexas.edu](https://useruploads.socratic.org/Vgpo3cORquDAhdHh9V7p_img527.png)
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

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