(B) The initial capacitance with air is given by $C = \frac{\epsilon_0 A}{d} = 1\,pF$.When the distance is doubled $(d' = 2d)$ and the space is filled

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(B) The initial capacitance with air is given by $C = \frac{\epsilon_0 A}{d} = 1\,pF$.When the distance is doubled $(d' = 2d)$ and the space is filled with a dielectric of constant $K$,the new capacitance $C'$ is given by $C' = \frac{K \epsilon_0 A}{d'}$.Substituting $d' = 2d$ into the equation,we get $C' = \frac{K \epsilon_0 A}{2d} = \frac{K}{2} \left( \frac{\epsilon_0 A}{d} \right) = \frac{K}{2} C$.Given $C' = 2\,pF$ and $C = 1\,pF$,we have $2 = \frac{K}{2} \times 1$.Solving for $K$,we get $K = 2 \times 2 = 4$.

Class 12 Physics Electric Potential and Capacitance Effect of Dielectric Inside Capacitor

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The capacitance of a capacitor is $1\,pF$ when there is air between its plates. If the distance between the plates is doubled and the space between them is filled with wax,the new capacitance becomes $2\,pF$. What is the dielectric constant of the wax?

A
$2$
B
$4$
C
$6$
D
$8$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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The capacitance of a capacitor is $1\,pF$ when there is air between its plates. If the distance between the plates is doubled and the space between them is filled with wax,the new capacitance becomes $2\,pF$. What is the dielectric constant of the wax?

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Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$, then the potential difference across the other capacitor will become:

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