Two capacitors of capacitances 20.0 pF and 50.0 pF are connected in...

Given $C_1 = 20.0 \text{ pF}$ and $C_2 = 50.0 \text{ pF}$ connected in series with $V = 6.00 \text{ V}$.

In a series combination, the charge $Q$ on each capacitor is the same.

The equivalent capacitance is $C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{20.0 \times 50.0}{20.0 + 50.0} = \dfrac{100}{7} \text{ pF}$.

Total charge $Q = C_{eq} V = \dfrac{100}{7} \times 6.00 = \dfrac{600}{7} \text{ pC}$.

Potential difference across the $20.0 \text{ pF}$ capacitor is $V_1 = \dfrac{Q}{C_1} = \dfrac{600/7}{20.0} = \dfrac{30}{7} \approx 4.29 \text{ V}$.

Potential difference across the $50.0 \text{ pF}$ capacitor is $V_2 = \dfrac{Q}{C_2} = \dfrac{600/7}{50.0} = \dfrac{12}{7} \approx 1.71 \text{ V}$.

Alternatively, using the voltage divider rule:

$V_1 = V \left( \dfrac{C_2}{C_1 + C_2} \right) = 6.00 \times \dfrac{50.0}{70.0} \approx 4.29 \text{ V}$

$V_2 = V \left( \dfrac{C_1}{C_1 + C_2} \right) = 6.00 \times \dfrac{20.0}{70.0} \approx 1.71 \text{ V}$