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Question

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} \R

Vedclass · Accepted Answer

$Q > Q_0$

Vedclass · Answer

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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