A parallel-plate capacitor has capacitance $C_0$ in vacuum. A dielectric slab of dielectric constant $K = 4$ is inserted to completely fill the space between the plates. What is the new capacitance?

Inserting a dielectric of constant $K$ multiplies the capacitance by $K$. With $K = 4$, $C = 4 C_0$. A common slip is to think the dielectric *reduces* capacitance — but it does the opposite. It *reduces* the electric field between the plates (by polarising), which lets the same plates store $K$ times more charge for the same voltage.