A charged capacitor when filled with a dielectric K = 3 has charge Q₀, voltage V₀ and field E₀. If the dielectric is replaced with another one having K = 9, the new values of charge, voltage and electric field will be respectively

When charged capacitor is filled with a dielectric material of dielectric constant K, then charge remains constant, V becomes $\frac{1}{\mathrm{K}}$ times and E becomes $\frac{1}{\mathrm{K}}$ times.

So, new value of charge is Q₀.

New value of $\mathrm{V}=\frac{{\mathrm{V}}_0\times 3}{9}=\frac{{\mathrm{V}}_0}{3}$

New value of $\mathrm{E}=\frac{{\mathrm{E}}_0\times 3}{9}=\frac{{\mathrm{E}}_0}{3}$