Condenser A has a capacity of 15,μ F when it is filled with a medium of dielectric constant 15 . An

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2. Electric Potential and Capacitance

hard

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.....$V$

$400$

$800$

$1200$

$1600$

Solution

(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.$

Standard 12

Physics

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Solution

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Assertion : If the distance between parallel plates of a capacitor is halved and dielectric constant is three times, then the capacitance becomes $6\,times$.

Reason : Capacity of the capacitor does not depend upon the nature of the material.