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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance

Two parallel plate capacitors \(C_1\) and \(C_2\) each having capacitance of \(10 \mu F\) are individually charged by a \(100\,V\) \(D.C.\) source. Capacitor \(C _1\) is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor \(C _2\) is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor \(C_1\) is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be \(.........V.\) (Assuming Dielectric constant \(=10\) )

  1. A \(40\)
  2. B \(50\)
  3. C \(55\)
  4. D \(65\)
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(C) \(55\)

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