Capacitors

JEE Main level, numerical answer, medium

24 of 180

Question 24 of 180

A capacitor PP with capacitance 10×10−6 F10 \times 10^{-6} \mathrm{~F} is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor PP is connected across another capacitor QQ with capacitance 20×10−6 F20 \times 10^{-6} \mathrm{~F}. The charge on capacitor QQ when equilibrium is established will be α×10−5C\alpha \times 10^{-5} C (assume capacitor QQ does not have any charge initially), the value of α\alpha is ____\_\_\_\_。

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Answer: 4

Capacitor P (10 × 10⁻⁶ F) is charged to 6.0 V.
Initial charge: QP=10×10−6×6=60×10−6Q_P = 10 \times 10^{-6} \times 6 = 60 \times 10^{-6} C.
When connected to uncharged capacitor Q (20 × 10⁻⁶ F), charge redistributes until both reach same potential.
Final voltage: Vf=QtotalCtotal=60×10−630×10−6=2V_f = \frac{Q_{total}}{C_{total}} = \frac{60 \times 10^{-6}}{30 \times 10^{-6}} = 2 V.
Charge on Q: QQ=20×10−6×2=40×10−6Q_Q = 20 \times 10^{-6} \times 2 = 40 \times 10^{-6} C = 4×10−54 \times 10^{-5} C.
Thus α=4\alpha = 4.

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