A linear dielectric that completely fills the region between capacitor conductors increases the capacitance by the relative permittivity $\varepsilon_r$:

$$\boxed{C=\varepsilon_r C_0},$$

where $C_0$ is the capacitance with vacuum between the conductors.

What happens to charge, voltage, electric field, and stored energy depends on what the capacitor is connected to while the dielectric is inserted.

Isolated capacitor: free charge fixed

Suppose the capacitor is charged and then disconnected from any source before the dielectric is inserted. No free charge can enter or leave, so

$$\boxed{Q=\text{constant}}.$$

Since

$$V=\frac{Q}{C},$$

increasing the capacitance by $\varepsilon_r$ gives

$$\boxed{V=\frac{V_0}{\varepsilon_r}}.$$

For a parallel-plate geometry with fixed plate separation $d$,

$$E=\frac{V}{d},$$

so

$$\boxed{E=\frac{E_0}{\varepsilon_r}}.$$

The dielectric polarization produces bound charges whose field partially opposes the field of the free plate charges, reducing the net field for the same $Q$.

The stored energy at fixed charge is

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

therefore

$$\boxed{U=\frac{U_0}{\varepsilon_r}}.$$

The field energy decreases. If the dielectric is pulled into the capacitor by electrostatic forces, part of that decrease can appear as mechanical work or other energy transfers.

Capacitor connected to an ideal voltage source: voltage fixed

Now suppose the capacitor remains connected to an ideal voltage source during insertion. The source maintains

$$\boxed{V=\text{constant}}.$$

Because

$$Q=CV,$$

the larger capacitance requires more free charge on the plates:

$$\boxed{Q=\varepsilon_r Q_0}.$$

For a fixed parallel-plate separation,

$$E=\frac{V}{d},$$

so the macroscopic field between ideal plates remains

$$\boxed{E=E_0}.$$

The voltage source supplies the additional free charge needed to maintain that field in the presence of polarization.

The capacitor field energy is now

$$U=\frac12CV^2,$$

so

$$\boxed{U=\varepsilon_r U_0}.$$

The energy stored in the capacitor field increases, but this does not mean energy was created. The voltage source transfers energy and charge into the capacitor-dielectric system during the process.

Comparison

For full insertion of a linear dielectric by factor $\varepsilon_r$:

The same dielectric therefore produces different observable changes depending on the external constraint.

Worked example

A vacuum capacitor has

$$C_0=100,\mathrm{pF}$$

and is initially charged to

$$V_0=200,\mathrm V.$$

Its initial charge is

$$Q_0=C_0V_0=(100\times10^{-12})(200) =20,\mathrm{nC}.$$

Insert a dielectric with

$$\varepsilon_r=4.$$

Then

$$C=400,\mathrm{pF}.$$

If the capacitor was disconnected first, $Q=20,\mathrm{nC}$ remains fixed and

$$V=\frac{200}{4}=\boxed{50,\mathrm V}.$$

If it stays connected to the $200,\mathrm V$ source instead, the voltage remains fixed and

$$Q=(400,\mathrm{pF})(200,\mathrm V)=\boxed{80,\mathrm{nC}}.$$

The dielectric property is the same in both experiments; the different outcomes come from whether charge or voltage is externally constrained.