Question

The capacitors in the figure are initially uncharged and are connected as in the diagram with switch $S$ open. The potential of point b after switch $S$ is closed and the amount of charge that flowed through the switch when it was closed are

• A. $66.7V,300\mu C$
• B. $100V,400\mu C$
• C. $100V,300\mu C$
• D. $133.3V,0\mu C$

Answer 1

The correct option is C $100V,300\mu C$

When the switch $S$ is open in both left and right branches, $6\mu F$ and $3\mu F$ are connected in series, so their effective capacitance is $\left(\frac{6\times 3}{6+3}\right)\mu F=2\mu F$
Thus, charge on each capacitor
($q=cv$, voltage across each capacitor is $200V$)
$=400\mu C$
Now, after closing key, pair of 6 𝜇𝐹 and 3 𝜇𝐹 are connected to each other in parallel in both upper and lower segments of circuit. So, equivalent capacitor in both segments will be equal i.e. $(6\mu F+3\mu F=9\mu F)$. Thus, potential difference 200 V will get distributed equally in two channel. Hence potential across each capacitor will be $100V$.
The charge distribution is shown below
Applying $q=cv$ for each capacitor on left side of switch
For $6\mu F$ ,
$q_1=\left(6\mu F\right)(100V$)
$=600\mu C$
Similarly , for $3\mu F$,
$q_4=\left(3\mu F\right)\left(100V\right)$
$=300\mu C$
Charge on lower plate of $600\mu F=-q_1=-600\mu C$
change in the charge on lower plate $\Delta q_1=-q_1-(-q)$
$=-600\mu C+400\mu C$
$=-200\mu C$
(change in charge of upper plate of $3\mu F,\Delta q{)}_4=q_4-q$
$=300-400$
$=-100\mu C$
So, charge flown through
the switch $=\Delta q_1+\Delta q_4$
$=-(100+200)\mu C$
$=-300\mu C$
Which is same as $300\mu C$ flow in opposite direction.
Hence, correct option is (B).