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There is an electric field between the plates because there is a potential difference between the capacitor plates. Namely, one face of the capacitor has charge $+Q$, and the other face has charge $-Q$. The voltage will be given by

$$ V(x) = \frac{Q}{C} = -E \ x . $$

The way to show this is to first use Gauss's law with the "pillbox". Each plate gives a contribution of

$$ E_{plate} = \frac{\sigma}{2\epsilon_0} $$

so that when you add them you get that

$$ E_{total} = \frac{\sigma}{\epsilon_0} $$

where $\sigma = \frac{Q}{A}$ where $Q$ is the charge induces on the capacitor and $A$ is the area of the plate.

To relate this to your question of $\nabla \cdot\mathbf{E}$, note that this is just the differential form of Gauss's law. Namely we have that

$$\iint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{1}{\epsilon_0} \iiint_V \nabla \cdot E \ dV = \frac{1}{\epsilon_0}\iiint_V \rho(\mathbf{r}) \ dV \equiv \frac{Q_{nec}}{\epsilon_0} $$

So while it is true that there is no charge density in between the plates, that doesn't cover the whole space. You need to consider the boundaries as well (namely the faces of the plates). In particular, for a parallel plate capacitor your density integral reduces to

$$ \iiint_V \rho(x) dV = \sigma A \int_{-\infty}^\infty dx\ \delta(x) + \delta(x-d)$$

if one plate is located at $\mathbf{x=0}$ and the other at $\mathbf{x =d}$.

I could go through the derivation of the electric field of a parallel plate capacitor, but these are easier to find than a parking space in Arkansas.

There is an electric field between the plates because there is a potential difference between the capacitor plates. Namely, one face of the capacitor has charge $+Q$, and the other face has charge $-Q$. The voltage will be given by

$$ V(x) = \frac{Q}{C} = -E \ x . $$

The way to show this is to first use Gauss's law with the "pillbox". Each plate gives a contribution of

$$ E_{plate} = \frac{\sigma}{2\epsilon_0} $$

so that when you add them you get that

$$ E_{total} = \frac{\sigma}{\epsilon_0} $$

where $\sigma = \frac{Q}{A}$ where $Q$ is the charge induced on the capacitor and $A$ is the area of the plate.

To relate this to your question of $\nabla \cdot\mathbf{E}$, note that this is just the differential form of Gauss's law. Namely we have that

$$\iint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{1}{\epsilon_0} \iiint_V \nabla \cdot E \ dV = \frac{1}{\epsilon_0}\iiint_V \rho(\mathbf{r}) \ dV \equiv \frac{Q_{nec}}{\epsilon_0} $$

So while it is true that there is no charge density in between the plates, that doesn't cover the whole space. You need to consider the boundaries as well (namely the faces of the plates). In particular, for a parallel plate capacitor your density integral reduces to

$$ \iiint_V \rho(x) dV = \sigma A \int_{-\infty}^\infty dx\ \delta(x) + \delta(x-d)$$

if one plate is located at $\mathbf{x=0}$ and the other at $\mathbf{x =d}$.

I could go through the derivation of the electric field of a parallel plate capacitor, but these are easier to find than a parking space in Arkansas.

There is an electric field between the plates because there is a potential difference between the capacitor plates. Namely, one face of the capacitor has charge $+Q$, and the other face has charge $-Q$. The voltage will be given by

$$ V(x) = \frac{Q}{C} = -E \ x . $$

The way to show this is to first use Gauss's law with the "pillbox". Each plate gives a contribution of

$$ E_{plate} = \frac{\sigma}{2\epsilon_0} $$

so that when you add them you get that

$$ E_{total} = \frac{\sigma}{\epsilon_0} $$

where $\sigma = \frac{Q}{A}$ where $Q$ is the charge induces on the capacitor and $A$ is the area of the plate.

To relate this to your question of $\nabla \cdot\mathbf{E}$, note that this is just the differential form of Gauss's law. Namely we have that

$$\iint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{1}{\epsilon_0} \iiint_V \nabla \cdot E \ dV = \frac{1}{\epsilon_0}\iiint_V \rho(\mathbf{r}) \ dV \equiv \frac{Q_{nec}}{\epsilon_0} $$

So while it is true that there is no charge density in between the plates, that doesn't cover the whole space. You need to consider the boundaries as well (namely the faces of the plates). In particular, for a parallel plate capacitor the charge density will be

$$ \rho(x) \propto \delta(\mathbf{x}) + \delta(\mathbf{x-d})$$

if one plate is located at $\mathbf{x=0}$ and the other at $\mathbf{x =d}$.

I could go through the derivation of the electric field of a parallel plate capacitor, but these are easier to find than a parking space in Arkansas.

There is an electric field between the plates because there is a potential difference between the capacitor plates. Namely, one face of the capacitor has charge $+Q$, and the other face has charge $-Q$. The voltage will be given by

$$ V(x) = \frac{Q}{C} = -E \ x . $$

The way to show this is to first use Gauss's law with the "pillbox". Each plate gives a contribution of

$$ E_{plate} = \frac{\sigma}{2\epsilon_0} $$

so that when you add them you get that

$$ E_{total} = \frac{\sigma}{\epsilon_0} $$

where $\sigma = \frac{Q}{A}$ where $Q$ is the charge induces on the capacitor and $A$ is the area of the plate.

To relate this to your question of $\nabla \cdot\mathbf{E}$, note that this is just the differential form of Gauss's law. Namely we have that

$$\iint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{1}{\epsilon_0} \iiint_V \nabla \cdot E \ dV = \frac{1}{\epsilon_0}\iiint_V \rho(\mathbf{r}) \ dV \equiv \frac{Q_{nec}}{\epsilon_0} $$

So while it is true that there is no charge density in between the plates, that doesn't cover the whole space. You need to consider the boundaries as well (namely the faces of the plates). In particular, for a parallel plate capacitor your density integral reduces to

$$ \iiint_V \rho(x) dV = \sigma A \int_{-\infty}^\infty dx\ \delta(x) + \delta(x-d)$$

if one plate is located at $\mathbf{x=0}$ and the other at $\mathbf{x =d}$.

I could go through the derivation of the electric field of a parallel plate capacitor, but these are easier to find than a parking space in Arkansas.

There is an electric field between the plates because there is a potential difference between the capacitor plates. Namely, one face of the capacitor has charge $+Q$, and the other face has charge $-Q$. The voltage will be given by

$$ V(x) = \frac{Q}{C} = -E \ x . $$

The way to show this is to first use Gauss's law with the "pillbox". Each plate gives a contribution of

$$ E_{plate} = \frac{\sigma}{2\epsilon_0} $$

so that when you add them you get that

$$ E_{total} = \frac{\sigma}{\epsilon_0} $$

where $\sigma = \frac{Q}{A}$ where $Q$ is the charge induces on the capacitor and $A$ is the area of the plate.

To relate this to your question of $\nabla \cdot\mathbf{E}$, note that this is just the differential form of Gauss's law. Namely we have that

$$\iint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{1}{\epsilon_0} \iiint_V \nabla \cdot E \ dV = \frac{1}{\epsilon_0}\iiint_V \rho(\mathbf{r}) \ dV \equiv \frac{Q_{nec}}{\epsilon_0} $$

I could go through the derivation of the electric field of a parallel plate capacitor, but these are easier to find than a parking space in Arkansas.

There is an electric field between the plates because there is a potential difference between the capacitor plates. Namely, one face of the capacitor has charge $+Q$, and the other face has charge $-Q$. The voltage will be given by

$$ V(x) = \frac{Q}{C} = -E \ x . $$

The way to show this is to first use Gauss's law with the "pillbox". Each plate gives a contribution of

$$ E_{plate} = \frac{\sigma}{2\epsilon_0} $$

so that when you add them you get that

$$ E_{total} = \frac{\sigma}{\epsilon_0} $$

where $\sigma = \frac{Q}{A}$ where $Q$ is the charge induces on the capacitor and $A$ is the area of the plate.

To relate this to your question of $\nabla \cdot\mathbf{E}$, note that this is just the differential form of Gauss's law. Namely we have that

$$\iint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{1}{\epsilon_0} \iiint_V \nabla \cdot E \ dV = \frac{1}{\epsilon_0}\iiint_V \rho(\mathbf{r}) \ dV \equiv \frac{Q_{nec}}{\epsilon_0} $$

So while it is true that there is no charge density in between the plates, that doesn't cover the whole space. You need to consider the boundaries as well (namely the faces of the plates). In particular, for a parallel plate capacitor the charge density will be

$$ \rho(x) \propto \delta(\mathbf{x}) + \delta(\mathbf{x-d})$$

if one plate is located at $\mathbf{x=0}$ and the other at $\mathbf{x =d}$.

I could go through the derivation of the electric field of a parallel plate capacitor, but these are easier to find than a parking space in Arkansas.