$A$ capacitor has some dielectric between its plates and the capacitor is connected to a $D.C.$ source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance,the energy stored in it,electric field,charge stored,and the voltage will increase,decrease,or remain constant.

(A) $1$. Capacitance $(C)$: The capacitance of a capacitor with a dielectric constant $K$ is given by $C = \frac{K \epsilon_0 A}{d}$. Since $K > 1$,removing the dielectric slab (replacing it with air,$K=1$) causes the capacitance $C$ to decrease.
$2$. Charge $(q)$: Since the battery is disconnected before the dielectric is removed,the capacitor is isolated. Therefore,the charge $q$ stored on the plates remains constant.
$3$. Energy stored $(U)$: The energy stored is given by $U = \frac{q^2}{2C}$. Since $q$ is constant and $C$ decreases,$U$ must increase.
$4$. Voltage $(V)$: From the relation $q = CV$,we have $V = \frac{q}{C}$. Since $q$ is constant and $C$ decreases,the voltage $V$ across the plates must increase.
$5$. Electric field $(E)$: The electric field between the plates is given by $E = \frac{V}{d}$. Since $V$ increases and $d$ remains constant,the electric field $E$ must increase.