A

Step 1: Understand the initial conditions of the parallel-plate capacitor. When a capacitor is charged by a battery, its initial energy (U_initial) can be expressed as:
$$ U_{initial} = \frac{1}{2} C V^2 $$
where C is the capacitance of the capacitor given by:
$$ C = \frac{\varepsilon_0 A}{d} $$
Thus,
$$ U_{initial} = \frac{1}{2} \left(\frac{\varepsilon_0 A}{d}\right) V^2 $$
Step 2: Upon disconnecting the battery, the charge (Q) on the capacitor remains constant and is given by:
$$ Q = C V = \frac{\varepsilon_0 A}{d} V $$
Step 3: When the dielectric slab of constant 2k is inserted, the new capacitance (C_new) becomes:
$$ C_{new} = k C = 2k \left(\frac{\varepsilon_0 A}{d}\right) = \frac{2k \varepsilon_0 A}{d} $$
Step 4: After inserting the dielectric slab, the voltage across the capacitor (V_new) can be found using the relation:
$$ V_{new} = \frac{Q}{C_{new}} = \frac{\frac{\varepsilon_0 A}{d} V}{\frac{2k \varepsilon_0 A}{d}} = \frac{V}{2k} $$
Step 5: The new potential energy (U_new) of the capacitor is:
$$ U_{new} = \frac{1}{2} C_{new} V_{new}^2 = \frac{1}{2} \left(\frac{2k \varepsilon_0 A}{d}\right) \left(\frac{V}{2k}\right)^2 $$
This simplifies to:
$$ U_{new} = \frac{1}{2} \left(\frac{2k \varepsilon_0 A}{d}\right) \left(\frac{V^2}{4k^2}\right) = \frac{\varepsilon_0 A V^2}{4d} $$
Step 6: The change in potential energy (ΔU) is then given by:
$$ \Delta U = U_{new} - U_{initial} $$
Substituting the values:
$$ \Delta U = \frac{\varepsilon_0 A V^2}{4d} - \frac{1}{2} \left(\frac{\varepsilon_0 A}{d}\right) V^2 $$
Simplifying gives:
$$ \Delta U = \frac{\varepsilon_0 A V^2}{4d} - \frac{2 \varepsilon_0 A V^2}{4d} = -\frac{\varepsilon_0 A V^2}{4d} $$
Conclusion: The change in potential energy is negative, indicating that the potential energy of the system decreases. Therefore, the final expression shows the change in potential energy due to the insertion of the dielectric slab.