Question

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

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

Answer 1

The correct option is C $5\mu C$ from b to a

When switch was open as shown in the figure 1, the equivalent capacitance across path PaQ is $C_{eq}=\frac{2\times 3}{2+3}\times {10}^{-6}F=1.5\mu F$.

Similarly the equivalent capacitance across path PbQ is also $C_{eq}=1.5\mu F$

When switch is closed the charge distribution across each capacitor is as shown in the figure 2.

Thus taking potential at the negative terminal of capacitor to be zero and at the switch to be $V_0$ We can write the charges across the capacitors as follows,

For path PaQ,

$Q_1=(10-V_0)\times 2\mu F...\left(i\right)$

$Q_2=(10-V_0)\times 3\mu F...\left(ii\right)$

For Path PbQ

$Q_1=V_0\times 2\mu F...\left(iii\right)$

$Q_2=V_0\times 3\mu F...\left(iv\right)$

Using equation (iii) in equation (i) we get $V_0=5V$

From equation (iii) and (iv) we get charges $Q_1\&Q_2$ as

$Q_1=10\mu C\&Q_2=15\mu C$

Thus charge flowing across switch is from point b towards point a

The net charge flowing is $15\mu C-10\mu C=5\mu C$