Question

In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch \[{{S}_{1}}\] is pressed first to fully charge the capacitor \[{{C}_{1}}\] and then released. The switch \[{{S}_{2}}\] is then pressed to charge the capacitor \[{{C}_{2}}\]. After some time, \[{{S}_{2}}\] is released and then \[{{S}_{3}}\] is pressed. After some time:
            

A. The charge on the upper plate of \[{{C}_{1}}\] is \[2C{{V}_{o}}\]
B. The charge on the upper plate of \[{{C}_{1}}\] is \[C{{V}_{o}}\]
C. The charge on the upper plate of \[{{C}_{1}}\] is 0
D. The charge on the upper plate of \[{{C}_{2}}\] is \[-C{{V}_{o}}\]

Answer 1

Hint: In this question we are given three different conditions for the circuit i.e. when the switches are pressed to complete the circuit and then released after some time. When we look at the options, we can say that we have been asked to calculate the charge on the capacitors at the end of the three conditions. Therefore, we will calculate the charge on capacitors for every condition and at the check which option best fits our calculated option.

Complete answer:
As shown in the figure, when the switch \[{{S}_{1}}\] is pressed the charge will flow through the circuit. As the other two switches are off the charge will only flow in the first part of the circuit as shown below.

Now, we know that charge is given by
\[Q=CV\]
It is given that capacitance of each capacitor is C
Therefore, after substituting values
The charge on first capacitor \[{{C}_{1}}\]is given by,
\[Q=2C{{V}_{o}}\]
Now, in the second condition the switch \[{{S}_{2}}\] is closed and charge flows in circuit as shown below.

Now, we know that the capacitors will be equally charged when the switch \[{{S}_{2}}\] is pressed. Therefore, half the charge of the first capacitor \[{{C}_{1}}\] will be acquired by the second capacitor \[{{C}_{2}}\]. Therefore, the charge on capacitor will be half the original charge
Therefore,
The charge on capacitor \[{{C}_{1}}\] will be \[Q=C{{V}_{o}}\];
 Also charge on capacitor \[{{C}_{2}}\]will be \[Q=C{{V}_{o}}\];
Similarly, at the third when only switch \[{{S}_{3}}\] is on (as shown in the figure below), the charge on capacitor \[{{C}_{2}}\] will be

\[Q=C{{V}_{o}}+C{{V}_{o}}=2C{{V}_{o}}\]
Therefore, at the end of the conditions we can say that,
Charge on capacitor \[{{C}_{1}}\] is \[Q=C{{V}_{o}}\]
And charge on capacitor \[{{C}_{2}}\] is \[Q=2C{{V}_{o}}\]

Therefore, the correct answer is option B.

Note:
Capacitor is an electronic device which stores electrical energy in the electric field. The capacitor consists of two parallel plates with opposite charge separated by a dielectric insulator. The charge Q stored in the capacitor is given as the product of capacitance in Farads and the voltage across it in Volts.