The capacitance of a parallel plate capacitor is $5\, \mu F$ . When a glass slab of thickness equal to the separation between the plates is introduced between the plates, the potential difference reduces to $1/8$ of the original value. The dielectric constant of glass is
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$\frac{V}{{{V_0}}} = \frac{{{C_0}}}{C} = \frac{1}{K} = \frac{1}{8}$
$\therefore $ $K = 8$
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