Condenser A has a capacity of \(15 \mu\text{F}\) when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity \(1 \mu\text{F}\) with air between the plates. Both are charged separately by a battery of \(100\text{ V}\). After charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is :
Solution:
Charge on A is \(15 \mu\text{F} \times 100\text{ V} = 1500 \mu\text{C}\) and on B is \(1 \mu\text{F} \times 100\text{ V} = 100 \mu\text{C}\). When dielectric of A is removed, its capacitance becomes \(1 \mu\text{F}\). The common potential is \(V_c = \frac{Q_{\text{total}}}{C_{\text{total}}} = \frac{1500 + 100}{1 + 1} = 800\text{ V}\).
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