JEE Physics 2026 Question with Solution

Given: Capacitor $P$ has $C_P = 10\times10^{-6}\,F$ and is initially charged to $V = 6.0\,V$. Capacitor $Q$ has $C_Q = 20\times10^{-6}\,F$ and is initially uncharged.

Find: The value of $\alpha$ if the final charge on capacitor $Q$ is $\alpha \times 10^{-5}\,C$.

Step 1: Find initial charge on capacitor $P$.

$$Q_0 = C_P V$$ $$Q_0 = (10\times10^{-6})(6.0) = 6\times10^{-5}\,\text{C}$$

Step 2: Apply charge conservation. After connecting capacitors $P$ and $Q$, total charge is conserved. Let the common final potential be $V_f$.

$$Q_0 = (C_P + C_Q)V_f$$ $$6\times10^{-5} = (10\times10^{-6} + 20\times10^{-6})V_f$$ $$6\times10^{-5} = 30\times10^{-6}V_f$$ $$V_f = 2\,\text{V}$$

Step 3: Find charge on capacitor $Q$.

$$Q_Q = C_Q V_f$$ $$Q_Q = (20\times10^{-6})(2) = 4\times10^{-5}\,\text{C}$$

Step 4: Compare with the given form.

$$Q_Q = \alpha\times10^{-5}\,\text{C}$$ $$\alpha = 4$$

Therefore, the value of $\alpha$ is $4$.

Given: One capacitor is initially charged and the other is uncharged.

Find: The final charge on capacitor $Q$ in the form $\alpha \times 10^{-5}\,C$.

When two capacitors are connected together, they finally attain the same potential. Since the battery is disconnected, the total charge of the isolated system remains constant.

Initial total charge:

$$Q_{\text{total}} = 6\times10^{-5}\,\text{C}$$

Equivalent capacitance after connection:

$$C_P + C_Q = 10\times10^{-6} + 20\times10^{-6} = 30\times10^{-6}\,\text{F}$$

Common final potential:

$$V_f = \frac{Q_{\text{total}}}{C_P + C_Q} = \frac{6\times10^{-5}}{30\times10^{-6}} = 2\,\text{V}$$

Charge on capacitor $Q$:

$$Q_Q = C_QV_f = (20\times10^{-6})(2) = 4\times10^{-5}\,\text{C}$$

Thus, in the form $\alpha \times 10^{-5}\,C$, the value is $\alpha = 4$.