A parallel place capacitor is charged by connecting a \(2 \mathrm{~V}\) battery across it. It is then disconnected from the battery and a glass slab is introduced between plates. Which of the following pairs of quantities decrease?

When a charged capacitor is disconnected from the battery and a glass slab (dielectric material) is introduced between the plates, then charge (Q), capacitance \((C)\), electric potential \((V)\) and stored energy \((U)\) is given as
\(Q=Q_0, C=K C_0, V=\frac{V_0}{K} \text { and } U=\frac{U_0}{K}\)
where, \(K=\) dielectric constant
\(V =\text {electric potential} \)
\( U =\text {potential energy} \)
\(V_0 =\text {initial potential of charged capacitor} \)
\( U_0 =\text {Initial energy of charged capacitor}\)
From above, we see that \(V\) and \(U\) decreases after introducing glass plate of dielectric constant \(k\) whereas capacitance increases and charge remains same because battery is disconnected after introducing dielectric material.