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The effect of inserting a dielectric (constant K) depends critically on whether the battery remains connected. This is one of the most frequently tested topics in JEE electrostatics.

Battery Connected (V constant): The battery maintains the voltage V across the capacitor. Inserting a dielectric increases C by factor K. Since V is fixed: Q = CV increases by K (battery supplies extra charge). E = $\frac{V}{d}$ remains unchanged (V and d unchanged). Energy U = \frac{1}{2}$$CV^2 increases by K (battery provides the energy). The dielectric is pulled in because the system energy increases at the battery's expense.

Battery Disconnected (Q constant): The charge Q on the plates cannot change (no current path). C increases by K. V = $\frac{Q}{C}$ decreases by K. E = $\frac{V}{d}$ decreases by K (equivalently, the polarization field opposes the free charge field). Energy U = Q^$\frac{2}{2C}$ decreases by K. The energy decrease equals the work done by the electric field in pulling the dielectric in.

In both cases, the dielectric is attracted into the capacitor — but for different reasons. With battery: system seeks maximum C (and battery compensates). Without battery: system seeks minimum energy (energy decreases as dielectric enters).

Force on a partially inserted dielectric (battery connected, plate width w): F = $\frac{1}{2}$(K-1)$epsilon_0$$\frac{w}{d}$$V^2$, constant and independent of insertion depth. Without battery: F = $\frac{1}{2}$(K-1)$epsilon_0$$\frac{w}{d}$$V^2$ where V varies with insertion depth — more complex calculation.

The distinction between these two scenarios is the single most important conceptual check in capacitor-dielectric problems.