Question #78627

1 Answer
Jan 18, 2017

Answer:

(A)

Explanation:

farside.ph.utexas.edu
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#