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Question

(B) The initial capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.When the distance between the plates is halved $(d'

Vedclass · Accepted Answer

$1.5$

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(B) The initial capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.When the distance between the plates is halved $(d' = d/2)$ and the space is filled with a dielectric medium of constant $K$,the new capacitance $C'$ is given by $C' = \frac{K \epsilon_0 A}{d'}$.Substituting $d' = d/2$ into the equation,we get $C' = \frac{K \epsilon_0 A}{d/2} = \frac{2K \epsilon_0 A}{d}$.Given that the new capacitance $C' = 3C$,we equate the expressions:$3 \left( \frac{\epsilon_0 A}{d} \right) = \frac{2K \epsilon_0 A}{d}$.Canceling $\frac{\epsilon_0 A}{d}$ from both sides,we get $3 = 2K$.Therefore,$K = \frac{3}{2} = 1.5$.

$A$ parallel plate capacitor has a capacitance $C$. The distance between the two plates is halved and the space between the plates is filled with a dielectric medium. If the new capacitance is $3C$,what is the dielectric constant of the medium?

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