A capacitor P with capacitance 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 . The charge on capacitor Q when equilibrium is estab

Question

A capacitor P with capacitance $10 \times 10^{- 6} F$ 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 $20 \times 10^{- 6} F$ . The charge on capacitor Q when equilibrium is established will be $α \times 10^{- 5} C$ .

(assume capacitor Q does not have any charge initially), the value of $α$ is __________. [2026]

Smart Achievers · Accepted Answer

(4)

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} = \frac{10^{- 5} \times 6 + 0}{3 \times 10^{- 5}}$

$V = 2 volt$

$Q_{2} = C_{2} V = 2 \times 10^{- 5} \times 2 = 4 \times 10^{- 5} C$

Solution

Ans.
(4)

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} = \frac{10^{- 5} \times 6 + 0}{3 \times 10^{- 5}}$

$V = 2 volt$

$Q_{2} = C_{2} V = 2 \times 10^{- 5} \times 2 = 4 \times 10^{- 5} C$

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Solution

Ans. (4) V=C1V1+C2V2C1+C2=10-5×6+03×10-5 V=2 volt Q2=C2V=2×10-5×2=4×10-5 C

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Ans.
(4)

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} = \frac{10^{- 5} \times 6 + 0}{3 \times 10^{- 5}}$

$V = 2 volt$

$Q_{2} = C_{2} V = 2 \times 10^{- 5} \times 2 = 4 \times 10^{- 5} C$