Two parallel plate capacitors C₁ and C₂, each having a capacitance of 10 μ F, are individually charged by a 100 V D.C. source. Capacitor C₁ is kept connected to the source and a dielectric slab is inserted between its plates. Capacitor C₂ is disconnected from the source and then a dielectric slab i…

Question

Two parallel plate capacitors C₁ and C₂, each having a capacitance of 10 μ F, are individually charged by a 100 V D.C. source. Capacitor C₁ is kept connected to the source and a dielectric slab is inserted between its plates. Capacitor C₂ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor C₁ 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)

Accepted Answer

55. C₁ stays across the source, so its potential is held at 100 V while its capacitance rises to 10×10=100 μ F: Q₁=100×100=10000 μ C. C₂ is disconnected first, so its charge is frozen at Q₂=10×100=1000 μ C even though its capacitance also rises to 100 μ F. Joined in parallel: total charge Q=10000+1000=11000 μ C, total capacitance C=100+100=200 μ F. V=Q/C=(11000)/(200)=55 V.

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