An air-filled parallel plate capacitor has plate spacing 1.20 cm and capacitance 4.80 μF. If the plate spacing is doubled and the space is completely filled with a dielectric, making the capacitance 16.5 μF, what is the dielectric constant of the material?

What changes when you double the spacing and add a dielectric?

For a parallel-plate capacitor, the capacitance depends on plate area, plate separation, and the permittivity of what fills the gap. Doubling the separation makes the capacitance smaller, but inserting a dielectric makes it larger.

Use the parallel-plate capacitor formula

With air (approximately vacuum), $$C_0 = \frac{\varepsilon_0 A}{d_0}$$ After doubling the spacing to $2d_0$ and filling with a dielectric of constant $\kappa$, $$C' = \frac{\kappa \varepsilon_0 A}{2d_0}$$

Relate the new capacitance to the old one

Divide $C'$ by $C_0$ to cancel $\varepsilon_0A$: $$\frac{C'}{C_0} = \frac{\kappa \varepsilon_0 A/(2d_0)}{\varepsilon_0 A/d_0} = \frac{\kappa}{2}$$ So, $$C' = \frac{\kappa}{2}C_0 \quad\Rightarrow\quad \kappa = 2\frac{C'}{C_0}$$

Plug in the numbers

$$\kappa = 2\frac{16.5}{4.80} = \frac{33.0}{4.80} = 6.875 \approx 6.88$$

The dielectric constant is $\boxed{\kappa \approx 6.9}$.