A parallel-plate capacitor, filled with a dielectric of dielectric constant k, is charged to a potential $V_{0}$ . It is now disconnected from the cell and the slab is removed. If it now discharges, with time constant $τ$ , through a resistance, the potential difference across it will be $V_{0}$ after time

k τ

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τ I n k

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τ I n (1 - \frac{1}{k})

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τ I n (k - 1)

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Solution

The correct option is C $τ I n k$
When the slab is removed, the potential of the capacitor increases k times, i.e., it becomes $k V_{0}$ .
Potential for discharging, $V = k V_{0} e^{- t / τ}$
The time t for potential drop to $V_{0}$ ,
Thus, $V = V_{0} = k V_{0} e^{- t / τ} \Rightarrow t = τ l n k$

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A parallel-plate capacitor, filled with a dielectric of dielectric constant k, is charged to a potential V0. It is now disconnected from the cell and the slab is removed. If it now discharges, with time constant τ, through a resistance, the potential difference across it will be V0 after time

The correct option is C τInkWhen the slab is removed, the potential of the capacitor increases k times, i.e., it becomes kV0.Potential for discharging, V=kV0e−t/τ The time t for potential drop to V0, Thus, V=V0=kV0e−t/τ⇒t=τlnk