Two parallel plate capacitors and each having capacitance of are individually charged by a 100 V D.C. source. Capacitor is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor 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)
Answer: 55
Solution
The capacitors start alike, each charged to . For the slab is slipped in while the battery is still attached, so the voltage is held at and the capacitance rises tenfold to ; its charge becomes . For the battery is removed first, so the charge is frozen at its initial even as the slab raises its capacitance to . Once both are isolated and joined in parallel, the charges pool and the capacitances add, . The common potential settles at .
JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Integer answer