Question

In the following figure is shown a system of four capacitors connected across a $10\, V$ battery. Charge that will flow from switch $S$ when it is closed is

• A. 5 $\mu\,C$ from $b$ to $a$
• B. 20 $\mu\,C$ from $a$ to $b$
• C. 5 $\mu\,C$ from $a$ to $b$
• D. $zero$

Answer 1

Answer: (A)

Using Kirchoff's law $-\frac{q_{1}}{2}+\frac{q_{2}}{3}=0$
$q_{1}=\frac{2}{3} q_{2} ; 10-\frac{q_{1}}{2}-\frac{q_{2}}{3}=0$
$\Rightarrow 60=3 q_{1}+2 q_{2}$
$\Rightarrow 60=3\left(\frac{2}{3} q_{2}\right)+2 q_{2}$
$\Rightarrow 4 q_{2}=60$
$\Rightarrow q_{2}=15 \mu c$
$\Rightarrow q_{1}=10 \mu c$
Therefore, charge through the switch
$q_{1}-q_{2}=-5 \mu c(a \rightarrow b)=5 \mu c( b \rightarrow a)$