The capacitance of a parallel plate capacitor is 10 $μ$ F When a dielectric plate is introduced in between the plates, its potential becomes V/4 th of its original value. What is the value of the dieletric constant of the plate introduced?

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Detailed Solution

C = KC (where K is the dielectric constant).
$\begin{array}{l} V = Q / C \\ V^{'} = Q / C^{'} \\ V^{'} = V / 4 = Q / C^{'} = Q / KC = V / K \\ ∴ K = 4 \end{array}$

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