1. Initial charging of both capacitors

Each capacitor has $C=10\,\mu F$ and each is charged by a $100\,V$ source.

So initial charge on each is $Q=CV=10\,\mu F \times 100\,V=1000\,\mu C.$

2. Capacitor $C_1$: kept connected to battery while dielectric is inserted

Since $C_1$ remains connected to the $100\,V$ source, its voltage stays constant.

Dielectric constant: $k=10$

So new capacitance of $C_1$ becomes $C_1'=kC=10\times 10\,\mu F=100\,\mu F.$

Because voltage remains $100\,V$, $Q_1'=C_1'V=100\,\mu F\times 100\,V=10000\,\mu C.$

Thus after dielectric insertion:

• Capacitance of $C_1 = 100\,\mu F$
• Charge on $C_1 = 10000\,\mu C$
• Potential on $C_1 = 100\,V$

3. Capacitor $C_2$: disconnected from battery before dielectric insertion

Since $C_2$ is disconnected first, its charge remains constant when dielectric is inserted.

Initial charge on $C_2$ was $Q_2=1000\,\mu C.$

After inserting dielectric, capacitance becomes $C_2'=kC=100\,\mu F.$

Now its new potential is $V_2' = \frac{Q_2}{C_2'}=\frac{1000\,\mu C}{100\,\mu F}=10\,V.$

Thus after dielectric insertion:

• Capacitance of $C_2 = 100\,\mu F$
• Charge on $C_2 = 1000\,\mu C$
• Potential on $C_2 = 10\,V$

4. Now both capacitors are disconnected and connected in parallel

They are connected in parallel with like plates together, so total charge is conserved.

Total initial charge of the combination: $Q_{\text{total}}=Q_1'+Q_2=10000+1000=11000\,\mu C.$

Total capacitance in parallel: $C_{\text{total}}=100\,\mu F+100\,\mu F=200\,\mu F.$

Therefore common potential is $V_f=\frac{Q_{\text{total}}}{C_{\text{total}}}=\frac{11000\,\mu C}{200\,\mu F}=55\,V.$

5. Final Answer

$\boxed{55\,V}$

This matches the stored correct answer.