A capacitor P with capacitance 10 × 10⁻⁶ F is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor P is connected across another capacitor Q with capacitance 20 × 10⁻⁶ F. The charge on capacitor Q when equilibrium is established will be α × 10⁻⁵ C (assume capacitor Q does not have any charge initially), the value of α is ___

A capacitor P with capacitance 10 × 10⁻⁶ F is fully charged with a potential difference of 6.0 V

Q. A capacitor $P$ with capacitance $10\times10^{-6}\ \text{F}$ is fully charged with a potential difference of $6.0\ \text{V}$ and disconnected from the battery. The charged capacitor $P$ is connected across another capacitor $Q$ with capacitance $20\times10^{-6}\ \text{F}$. The charge on capacitor $Q$ when equilibrium is established will be $\alpha\times10^{-5}\ \text{C}$ (assume capacitor $Q$ does not have any charge initially), the value of $\alpha$ is ____.

Correct Answer: 4

Explanation

This is a standard charge sharing problem frequently asked in JEE Main, JEE Advanced and IIT JEE examinations.

Initial charge on capacitor $P$ is

$$ Q_{\text{initial}} = C_P V $$

$$ Q_{\text{initial}} = 10\times10^{-6} \times 6.0 = 60\times10^{-6}\ \text{C} $$

After connection, both capacitors come to the same final potential. Total charge remains conserved.

Equivalent capacitance of the system is

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

Final common potential is

$$ V_f = \frac{Q_{\text{initial}}}{C_{\text{eq}}} $$

$$ V_f = \frac{60\times10^{-6}}{30\times10^{-6}} = 2\ \text{V} $$

Charge on capacitor $Q$ is

$$ Q_Q = C_Q V_f = 20\times10^{-6} \times 2 = 40\times10^{-6}\ \text{C} $$

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

Hence, $\alpha = 4$.

Explanation

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