Question

A 3.0 𝜇𝐹 and a 6.0 𝜇𝐹 capacitors are connected

in parallel to a 6.00 𝑉 battery? What are the

charges of the capacitors?

Answer 1

The potential difference across each element in the parallel circuit is the same. Therefore, we can find the charges of the capacitors:

$$Q_1=C_1\cdot\Delta V=3.0\cdot10^{-6}\ F\cdot6\ V=18\ \mu C,$$ $$Q_2=C_2\cdot\Delta V=6.0\cdot10^{-6}\ F\cdot6\ V=36\ \mu C.$$