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

\frac{d}{3}

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\frac{5 d}{9}

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\frac{7 d}{9}

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d

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Solution

The correct option is C $\frac{7 d}{9}$

$C_{f} = \frac{ϵ_{0} A}{d - t + \frac{t}{K}}$
Initially,
$C = \frac{ϵ_{0} A}{d}$
$\Rightarrow C_{f} = \frac{C d}{d - t + \frac{t}{\infty}}$
$\Rightarrow C_{f} = \frac{C d}{d - t}$
$4.5 d - 4.5 t = d$
$4.5 t = 3.5 d$
$t = \frac{7 d}{9}$

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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

The correct option is C 7d9 Cf=ϵ0Ad−t+tK Initially, C=ϵ0Ad ⇒Cf=Cdd−t+t∞ ⇒Cf=Cdd−t 4.5d−4.5t=d 4.5t=3.5d t=7d9

The correct option is C $\frac{7 d}{9}$

$C_{f} = \frac{ϵ_{0} A}{d - t + \frac{t}{K}}$
Initially,
$C = \frac{ϵ_{0} A}{d}$
$\Rightarrow C_{f} = \frac{C d}{d - t + \frac{t}{\infty}}$
$\Rightarrow C_{f} = \frac{C d}{d - t}$
$4.5 d - 4.5 t = d$
$4.5 t = 3.5 d$
$t = \frac{7 d}{9}$