electrostatics - Why do capacitors in a dielectric have the same charge as without the dielectric

If you imagine a dielectric of height $d$ and dielectric constant $\kappa$ placed evenly in a capacitor of height $D$. We define $\Delta x = \frac{D-d}{2}$.

The equivalent capacitance is now

$$\begin{align} \frac{1}{C_{eq}} &= 2 \frac{\Delta x }{\epsilon_0 A} + \frac{d}{\kappa \epsilon_0 A} \\ &= \frac{D-d}{\epsilon_0 A} + \frac{d}{\kappa \epsilon_0 A} \\ &= \frac{\kappa(D-d)+d}{\kappa\epsilon_0 A} \end{align} $$

At this point

$$C_{eq} = \frac{\kappa\epsilon_0 A}{\kappa(D-d)+d}$$

Based on prior knowledge, it should be that the charge on the equivalent capacitor is the same as the charge on the original, non dielectric capacitor

In other words:

$q = CV = C_{eq}V$ and voltage hasn't changed.

But $CV \ne C_{eq}V$!

What am I missing?

Answer

My friend, you spoke too soon. If you find the new voltage,

$$V^\prime = \frac{q\kappa(D-d)x + qd}{\epsilon_0 A \kappa} = \frac{q}{\epsilon_0A\kappa}\bigg(\kappa(D-d)+d\bigg)$$

Where $$C_{eq} = \frac{\kappa \epsilon_0 A}{k(D-d)+d}$$

Multiply out $C_{eq}V^\prime$ everything cancels and you are left with:

$$q^\prime = C_{eq} V^\prime = q$$

Therefore $q$ is constant wrt introduction of dielectrics.

Clarification Regarding $\kappa$

In the derivation we implicitly assume that $E^\prime = \frac{E_0}{\kappa}$, the same $\kappa$ from $\kappa = \frac{C^\prime}{C}$. The textbook justifies this by writing that such is true. That we are dealing with the same kappa in both of the last two equations is quite startling – what is the motivation for making the assumption that $E^\prime = \frac{E_0}{\kappa}$.

Doing a little bit of contextualization work, we can postulate that $\kappa$ was originally defined as $$\kappa = \frac{q}{q-q^\prime}$$ where $q^\prime$ is the charge induced. With this being true, we go through a set of simultaneous steps beginning with

$$E_0= \frac{q}{\epsilon_0 A} \qquad E^\prime = \frac{q-q^\prime}{\epsilon_0 A} = \frac{q}{\epsilon_0 A \kappa } = \frac{E_0}{\kappa} $$

Leading to

$$V_0= \frac{qd}{\epsilon_0 A} \qquad V^\prime = \frac{qd}{\epsilon_0 A} \kappa$$

Plugging the expressions for $V_0$ and $V^\prime$ into the definition of capacitance $C = \frac{q}{V}$ we obtain that

$$\frac{C^\prime}{C} = \kappa$$

This last broad statement about capacitance emerges from the definition of $\kappa$ as it relates the the microscopic concern of induced charges on the dielectric.