The electric field in the dielectric slab is:

\frac{V}{K d}

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\frac{K V}{d}

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\frac{V}{d}

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\frac{K V}{t}

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Solution

The correct option is A $\frac{V}{K d}$
Initially, the electric filed is $E_{0} = \frac{Q}{A ϵ_{0}}$
also, $E_{0} = \frac{V}{d}$
As battery is disconnected so charge Q is constant.
After inserted the dielectric the electric field becomes $E^{'} = \frac{Q}{A K ϵ_{0}} = \frac{E_{0}}{K} = \frac{V}{K d}$

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The electric field in the dielectric slab is:

The correct option is A VKdInitially, the electric filed is E0=QAϵ0 also, E0=VdAs battery is disconnected so charge Q is constant.After inserted the dielectric the electric field becomes E′=QAKϵ0=E0K=VKd