Two capacitors of capacitances 4μF and 12μF respectively, are connected in series. The combination is then connected momentarily across a 200V battery. Find the charge and potential difference across each capacitor.

Given,
Capacitor $(C_1) = 4\mu F = 4 * 10^{-6}\;F$
Capacitor $(C_2) = 12\mu F = 12 * 10^{-6}\;F$
Potential $(V) = 200 \;V$
Charge $(q) = ?$
Potential Difference in each capacitor $V_1 \; \& \; V_2 = ?$

When the capacitors are connected in series, then the equivalent capacitance

$C = $ $\frac{C_1 \; * \; C_2}{C_1 \; + \; C_2} = \frac{ 4 \; * \; 10^{-6} \; * \;  12 \; * \; 10^{-6}}{4 \; * \; 10^{-6} \; + \;  12 \; * \; 10^{-6}}$ $ = 3 \; * \; 10^{-6}\;F$

Now, we have to find Charge $(q) = C\;*\;V = 3 \; * \; 10^{-6} \; * \; 200 = 6 \; * \; 10^{-4}\; C$

Again,
Potential across the first capacitor $V_1 = $ $\frac{q}{C_1} = \frac{6 * 10^{-4}}{4 * 10^{-6}}$ $= 150\;V$

Potential across the second capacitor $V_2 = $ $ \frac{q}{C_2} = \frac{6 * 10^{-4}}{12 * 10^{-6}}$ $= 50\;V$

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