Condenser $A$ has a capacity of $15\,\mu F$ when it is filled with a medium of dielectric constant $15$. Another condenser $B$ has a capacity of $1\,\mu F$ with air between the plates. Both are charged separately by a battery of $100\;V$. After charging, both are connected in parallel without the battery and the dielectric medium being removed. The common potential now is.....$V$
A$400$
B$800$
C$1200$
D$1600$
Diffcult
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B$800$
b (b) 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 $C_{eq}$ $ = 1 + 1 = 2\,\mu F$
So common potential $ = \frac{{(15 \times {{10}^{ - 4}}) + (1 \times {{10}^{ - 4}})}}{{2 \times {{10}^{ - 6}}}} = 800\,V.$
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