(1)
If it is charged to a voltage (potential) V, the charge on its plates is Q = CV.
Since the battery is disconnected after it is charged, the charge Q on its plates, and consequently the product CV, remain unchanged.
On removing the dielectric completely, its capacitance becomes from Eq. (1),
$
C^{\prime}=\frac{\varepsilon_0 A}{d}=\frac{1}{k} C
$
that is, its capacitance decreases by the factor $k$. Since $C^{\prime} V^{\prime}=C V$, its new voltage is $V ^{\prime}=\frac{C}{C^{\prime}} V = kV$
so that its voltage increases by the factor $k$. The stored potential energy, $U=\frac{1}{2} Q V$, so that $Q$ remaining constant, $U$ increases by the factor $k$. The electric field, $E=V / d$, so that $E$ also increases by a factor $k$.
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