A capacitor $$C_1 = 6\,\mu\text{F}$$, is charged to a potential difference of $$V_{0} = 5$$ using a 5 V battery. The battery is removed and another capacitor, $$C_2 = 12\,\mu\text{F}$$ is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges ($$q_{1}$$ and $$q_{2}$$ on the capacitors $$C_{1}$$ and $$C_{2}$$ when equilibrium condition is reached.

step 1: initial condition

only $$C_1$$​ is charged:

$$q_{\text{initial}}=C_1V=6\mu F\times5=30\mu C$$

$$C_2$$​ is uncharged

step 2: after switch is closed

the two capacitors are connected in parallel (like plates joined)

so:

• total charge is conserved = 30 μC
• final voltage across both is same

step 3: find final voltage

$$C_{\text{eq}}=C_1+C_2=6+12=18\mu F$$

$$V_f=\frac{Q_{\text{total}}}{C_{\text{eq}}}=\frac{30}{18}=\frac{5}{3}\text{ V}$$

step 4: final charges

$$q_1=C_1V_f=6\times\frac{5}{3}=10\mu C$$

$$q_2=C_2V_f=12\times\frac{5}{3}=20\mu C$$