Two capacitors of capacitances $4\mu F$ and $6\mu F$ respectively are joined in series with a battery of emf $60\; V$. The connections are broken and the like terminals of the capacitors are then joined. Find the final charge on each capacitor.

Given,
Capacitance of capacitors $(C_1) = 4\mu F$
Capacitance of capacitor $(C_2) = 6\mu F$
Potential $(V) = 60\;V$
Charge in each capacitor $= ?$

Firstly the combination is in series, then
Total Capacitance $(C) = $ $\frac{C_1\;*\;C_2}{C_1 \; + \; C_2} = \frac{4 \;* \;10^{-6}\;*\;6\;*\;10^{-6}}{4 \;*\;10^{-6} \; + \; 6\;*\;10^{-6}}$ $= 2.4\;*\;10^{-6}\;F$

In series combination charge in each capacitor is same, then

Charge $(q) = C\;*\;V = 60\;*\;2.4\;*\;10^{-6} = 1.44\;*\;10^{-4}\;C$

But Potential in each capacitor is,

$V_1 = $ $ \frac{q}{C_1} = \frac{1.44\;*\;10^{-4}}{4\;*\;10^{-6}}$ $ =36\;V $

$V_2 = $ $\frac{q}{C_2} = \frac{1.44\;*\;10^{-4}}{6\;*\;10^{-6}}$ $ =24\;V $

Now, when they are connected in parallel, the potential is the same. Then the common potential is $V$ given by

$V = $ $\frac{C_1V_1 \;+\; C_2 V_2}{C_1 \;+\; C_2} = \frac{(4 \;*\; 10^{-6}\;*\; 36 )\;+\; (6 \;*\; 10^{-6} \;*\; 24)}{4 \;*\; 10^{-6} \;+\; 6 \;*\; 10^{-6}}$ $= 28.8\;V$

In parallel combination,

The charge on the first capacitor,

$q_1 = C_1 * V = 4 * 10^{-6}*28.8 = 1.152 * 10^{-4}\;C$

The charge on the second capacitor,

$q_2 = C_2 * V = 6 * 10^{-6}*28.8 = 1.728 * 10^{-4}\;C$

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