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The introduction of a metal plate between the plates of a parallel plate capacitor increases its capacitance by 4.5 times. If d is the separation of the two plates of the capacitor, the thickness of the metal plate introduced is

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$\frac{7 d}{9}$

$\frac{5 d}{9}$

$\frac{d}{3}$

answer is C.

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Detailed Solution

Initial capacitance C = $\frac{ε_{0} A}{d}$ . When a metal plate of thickness t is introduced, the capacitance becomes

$C' = \frac{ε_{0} A}{(d - t)} . G i v e n C' = 4.5 C T h u s \frac{ε_{0} A}{(d - t)} = \frac{ε_{0} A}{d} \times \frac{9}{2} w h i c h g i v e s 9 (d - t) = 2 d o r t = \frac{7 d}{9}$

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