JEE Physics 2023 Question with Solution

Given: Two capacitors $C_1$ and $C_2$ each have capacitance $C = 10 \, \mu \text{F}$ and are charged by a $100 \, \text{V}$ source. The dielectric constant is $K = 10$.

Find: The common potential after both capacitors are disconnected and connected in parallel.

For capacitor $C_1$, it remains connected to the source while the dielectric is inserted, so its capacitance becomes $KC$ and its voltage remains $V$. Hence its charge becomes

$$Q_1 = KCV$$

For capacitor $C_2$, it is disconnected from the source before inserting the dielectric, so its charge remains unchanged:

$$Q_2 = CV$$

When the two capacitors are connected in parallel, total charge is conserved:

$$Q_{\text{total}} = Q_1 + Q_2 = KCV + CV = (K+1)CV$$

Their final total capacitance is

$$C_{\text{eq}} = KC + KC = 2KC$$

Therefore, the common potential is

$$V_f = \frac{Q_{\text{total}}}{C_{\text{eq}}} = \frac{(K+1)CV}{2KC} = \frac{K+1}{2K}V$$

Now substitute $K=10$ and initial $V=100 \, \text{V}$:

$$V_f = \frac{11}{20} \times 100 = 55 \, \text{V}$$

Therefore, the common potential is $55 \, \text{V}$.

Given: $Q_1 = KCV$ and $Q_2 = CV$ from the two cases of dielectric insertion.

Find: Final common voltage.

Since both capacitors finally have dielectric inserted, each has capacitance $KC$. On parallel connection,

$$q = \frac{Q_1 + Q_2}{2} = \frac{KCV + CV}{2} = \frac{K+1}{2}CV$$

This is the charge effectively associated with one branch after redistribution.

Now for one capacitor,

$$V_f = \frac{q}{KC} = \frac{\frac{K+1}{2}CV}{KC} = \frac{K+1}{2K}V$$

Putting $K=10$ and $V=100 \, \text{V}$,

$$V_f = \frac{11}{20} \times 100 = 55 \, \text{V}$$

Therefore, the correct numerical value is 55.