A dielectric multiplies C by kappa. The clamp does the rest

A dielectric has no free charge. Put it in a field and its molecules stretch and align, so the positive ends of every molecule shift slightly one way and the negative ends the other. In the bulk these displacements cancel neighbor against neighbor; at the two surfaces they do not, and a layer of bound surface charge $\pm\sigma_b$ is left behind.

That bound charge sets up a field opposing the applied one, so the net field inside the slab is reduced:

$$E_{\text{inside}} = \frac{E_0}{\kappa}, \qquad \sigma_b = \sigma_f\left(1 - \frac{1}{\kappa}\right),$$

where $\sigma_f$ is the free charge on the plates and $\kappa \geq 1$ is the dielectric constant. Reduced, not erased. For any real material $\kappa$ is finite, so $E_0/\kappa$ is finite.

Contrast it with a conducting slab in the same gap:

1. Conductor: free electrons move macroscopic distances, pile up until the interior field is exactly zero, and can be drained off through a wire. This is the $\kappa \to \infty$ limit.
2. Dielectric: charges are bound to their molecules and shift by less than an atomic diameter. The interior field is reduced by the factor $\kappa$, and the bound surface charge cannot be conducted anywhere; wire it up and nothing flows.

Push the field high enough and the material stops being a dielectric altogether: above its dielectric strength, roughly $3\times10^6$ V/m for air, the medium ionizes and conducts. Every real capacitor's voltage rating is that limit written in volts.

Filling the gap replaces $\varepsilon_0$ with $\kappa\varepsilon_0$ everywhere it appeared, so

$$C = \frac{\kappa\varepsilon_0 A}{d} = \kappa C_0.$$

That is the only unconditional statement. Now ask what the wiring holds fixed, and read down the matching column.

1. Battery attached: $V$ is clamped. $C \to \kappa C$, so $Q = CV \to \kappa Q$. The field $E = V/d$ is unchanged, because both the voltage and the gap are unchanged. Energy $U = \tfrac12 CV^2 \to \kappa U$: the battery pushes more charge in and the stored energy goes UP.
2. Disconnected: $Q$ is clamped. $C \to \kappa C$, so $V = Q/C \to V/\kappa$ and $E = V/d \to E/\kappa$. Energy $U = Q^2/2C \to U/\kappa$: the stored energy goes DOWN.

The two columns disagree about the sign of nearly every change, which is why memorizing one row is worse than useless. What is common to both: $C$ grows by $\kappa$, and the capacitor can now hold more charge at any given voltage, which is the entire reason dielectrics are in real capacitors at all. The second reason is mechanical: the slab keeps the plates from touching, so $d$ can be made tiny.

When the slab does not fill the gap, stop and ask which quantity the two regions are forced to share. That answer names the combination, and the combination is the whole method.

1. Slab fills part of the AREA, full thickness. The filled region and the empty region are side by side, both spanning plate to plate, so both have the same potential difference. That is parallel: $C = C_{\text{empty}} + C_{\text{filled}}$, each computed with its own area.
2. Slab fills part of the GAP THICKNESS, full area. The two regions are stacked, and the field lines pass through one and then the other, so they carry the same charge. That is series: $1/C = 1/C_1 + 1/C_2$, each computed with its own thickness.

Work the stacked case once, with a slab of constant $\kappa$ filling half the gap:

$$C_1 = \frac{\varepsilon_0 A}{d/2} = 2C_0, \qquad C_2 = \frac{\kappa\varepsilon_0 A}{d/2} = 2\kappa C_0,$$

$$\frac{1}{C} = \frac{1}{2C_0} + \frac{1}{2\kappa C_0} \quad\Longrightarrow\quad C = \frac{2\kappa}{1 + \kappa}\,C_0.$$

For $\kappa = 3$ that is $1.5\,C_0$. The tempting shortcut, an "effective" $\kappa_{\text{eff}} = (1 + \kappa)/2 = 2$, gives $2C_0$ and is simply a different number. Averaging $\kappa$ is not a weaker method; it is not a method.

It is worth noticing that for the half-AREA case the parallel result does come out as $C_0(1 + \kappa)/2$, which is numerically what the averaging shortcut would give. That coincidence is what keeps the shortcut alive. It fails the moment the geometry is stacked instead of side by side, or the split is anything other than half and half by area, so the decomposition is what to practice.

Hold a neutral dielectric slab at the mouth of a charged capacitor and let go. It is drawn IN. The mechanism is the fringing field at the edge of the plates: it polarizes the slab, and the induced bound charge nearest each plate is opposite to that plate's charge and closer to it than its partner, so the net force has a component pointing into the gap.

The energy argument gives the same answer and gives its size. At fixed $Q$ (isolated capacitor), inserting the slab raises $C$, and

$$U = \frac{Q^2}{2C}$$

falls. A system pulled toward lower energy is a system under a force in that direction, and

$$F_x = -\frac{dU}{dx}$$

with $x$ the inserted length gives the magnitude directly, since $C(x)$ for a partially inserted slab is the parallel combination of an inserted piece and an empty piece.

With the battery attached the force is still inward, but the bookkeeping needs a third term. Here $U = \tfrac12 CV^2$ increases as the slab enters, and the battery supplies $\Delta W_{\text{batt}} = V\Delta Q = \Delta Q\,V$, which is twice the increase in stored energy; the leftover half is the work done on the slab. Energy conservation with a source in the loop always needs the source's contribution on the books.

What is never right is "the slab is neutral, so nothing happens to it". Neutral is not the same as unpolarizable, and every attraction between a charged object and a neutral insulator in this course works this way.