Two parallel plate capacitors C₁ and C₂ each having 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 it 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 )

Solution

Physics sir · 4 steps

Battery attached fixes V, battery removed fixes Q: get each charge first, then let them share.

1. Capacitor C1. With the dielectric, C₁ = KC = 10 × 10 = 100 μF. It stays on the 100 V source, so the source pushes in extra charge: Q₁ = 100 × 100 = 10000 μC. Removing the source afterwards does not change this charge.
2. Capacitor C2. It is charged first: Q₂ = 10 × 100 = 1000 μC. Then it is cut off, so this charge is locked in. The slab raises C₂ to 100 μF, and its voltage drops to 1000/100 = 10 V.
3. Connect in parallel. Positive plates are joined to positive plates. Charge is conserved and both capacitors end at the same voltage, so divide the total charge by the total capacitance.

   V = (Q₁ + Q₂)/(C₁ + C₂) = (10000 + 1000)/(100 + 100) V = 11000/200 V

4. So. The common potential is 55 V.

The common slip: 10. Keeping C₁'s charge at 1000 μC, as if no extra charge flowed in while it was on the source, gives 2000/200 = 10 V instead of 55.

The idea to keep. First ask whether the battery is attached. Attached: V stays fixed, Q changes. Detached: Q stays fixed, V changes. In parallel, the common V is total charge over total capacitance.

Answer from the NTA final answer key. Solution by PrepWiser.