A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with the capacitor are given by $Q_0, V_0, E_0$ and $U_0$ respectively. A dielectric slab is introduced between plates of capacitor but battery is still in connection. The corresponding quantities now given by $Q, V, E$ and $U$ related to previous ones are
Medium
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As the battery till in connection so potential will remained unchanged. i,e $V=V_{0}$
$E_{0}=\frac{V_{0}}{d}$ and $E=\frac{V}{d}=\frac{V_{0}}{d} \Rightarrow E=E_{0}$ where $d=$ separation of plates.
Let initial capacitance $=C_{0}$
When dielectric slab with dielectric constant $\mathrm{K}$ is inserted, the capacitance becomes $C=K C_{0}$
Thus, $Q_{0}=C_{0} V_{0}$ and $Q=C V=K C_{0} V_{0}=K Q_{0} \Rightarrow Q>Q_{0}$ as $(K>1)$
$\mathrm{Also}, U_{0}=(1 / 2) C_{0} V_{0}^{2}$ and $U=(1 / 2) C V^{2}=(1 / 2) K C_{0} \cdot V_{0}=K U_{0}$
Hence, $U>U_{0}$
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