Two capacitors of capacitances 5μF and 15μF respectively are connected in series with a battery of 220V . Find the charge and potential across each capacitor.

Given,
Capacitor $(C_1) = 5\mu F = 5 * 10^{-6}\;F$
Capacitor $(C_2) = 15\mu F = 15 * 10^{-6}\;F$
Potential $(V) = 220 \;V$
Charge $(q) = ?$
Potential Difference in each capacitor $V_1 \; \& \; V_2 = ?$

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

$C = $ $\frac{C_1 \; * \; C_2}{C_1 \; + \; C_2} = \frac{ 5 \; * \; 10^{-6} \; * \;  15 \; * \; 10^{-6}}{5 \; * \; 10^{-6} \; + \;  15 \; * \; 10^{-6}}$ $ = 3.75 \; * \; 10^{-6}\;F$

Now, we have to find Charge $(q = q_1 = q_2) = C\;*\;V = 3.75 \; * \; 10^{-6} \; * \; 220 = 8.25 \; * \; 10^{-4}\; C$

Again,
Potential across the first capacitor $V_1 = $ $\frac{q}{C_1} = \frac{8.25 * 10^{-4}}{5 * 10^{-6}}$ $= 165\;V$

Potential across the second capacitor $V_2 = $ $ \frac{q}{C_2} = \frac{8.25 * 10^{-4}}{15 * 10^{-6}}$ $= 55\;V$

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