Condenser has a capacity of 15μF when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity of 1μF with air between the plates. Both are charged separately by a battery of 100V. After charging, both are connected in parallel without the battery and the dielectric medium being removed. The common potential now is

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400V

1200 V

1600 V

800V

answer is B.

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Detailed Solution

Charge on capacitor A is given by ${Q_1} = 15 \times {10^{ - 6}} \times 100 = 15 \times {10^{ - 4}}C$

Charge on capacitor B is given by ${Q_2} = 1 \times {10^{ - 6}} \times 100 = {10^{ - 4}}C$

Capacity of capacitor A after removing dielectric $= \frac{{15 \times {{10}^{ - 6}}}}{{15}} = 1\mu F$

Now when both capacitors are connected in parallel their equivalent capacitance will be ${C_{ea}} = 1 + 1 = 2\mu F$

So common potential $= \frac{{(15 \times {{10}^{ - 4}}) + (1 \times {{10}^{ - 4}})}}{{2 \times {{10}^{ - 6}}}} = 800V.$

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