Question 81 of 180

The distance between two plates of a capacitor is $d$ and its capacitance is ${C}_{1}$, when air is the medium between the plates. If a metal sheet of thickness $\dfrac{2d}{3}$ and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes ${C}_{2}$ . The ratio $\dfrac{{C}_{2}}{{C}_{1}}$ is

A$3:1$B$2:1$C$4:1$D$1:1$

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Answer: (A) $3:1$

The formula of capacitance is ${C}_{1}=\left(\dfrac{{\epsilon }_{0}A}{d}\right)$

The dielectric constant of metal sheet is ${K}_{metal}=\infty$
The new value of capacitance is
${C}_{2}=\dfrac{{\epsilon }_{0}A}{d-t+\dfrac{t}{{K}_{metal}}}=\dfrac{{\epsilon }_{0}A}{d-\dfrac{2d}{3}+0}=3{C}_{1}$
Clearly, the ratio,
$\dfrac{{C}_{2}}{{C}_{1}}=\dfrac{3{C}_{1}}{{C}_{1}}=3:1$