A parallel$-$plate capacitor has a capacitance of $2\mathrm{.}6F$ with air between the plates. The capacitor is connected to a $6\mathrm{.}7-$volt battery and charged. The battery is then removed. When a dielectric is placed between the plates, the capacitance changes to $9\mathrm{.}4F.$ What is the voltage between the plates when the dielectric is present?

Report your answer in units of volts. For example, if your calculations produce a value of $1\mathrm{.}23$ V $,$ enter your answer as $1\mathrm{.}23.$