Question

Consider three identical, ideal capacitors. The first capacitor is charged to a voltage V and thendisconnected from the battery. The other two capacitors, initially uncharged and connected in series,are then connected across the first. What is the final voltage across the first capacitor?

Answer 1

Let each capacitor has capacitance $C$

Given,first capacitor charged to $Q_1=CV$

Now, as other two capacitor is connected in series ,resultant capacitance will be

$$\frac{1}{C_{res}}=\frac{1}{C}+\frac{1}{C}\implies C_{res}=\frac{C}{2}$$

Since, charge in the capacitor circuit(Ideal) is conserved,thus

$$q_1+q_2=Q_1=CV$$

where, $q_1,q_2$ are charge on first and resultant capacitor respectively.

Now, charge flowing will stop when both capacitor will have same potential,thus

$$\frac{q_1}{C}=\frac{q_2}{C/2}\implies q_1=2q_2$$

Thus, we get

$$q_1=\frac{2CV}{3}\implies \frac{q_1}{C}=V'=\frac{2}{3}V$$

Therefore,final voltage across first capacitor is $\frac{2}{3}V$ .