91Ó°ÊÓ

The formulae required

$$\begin{gathered} \mathrm{Time}\left(\mathrm{t}\right)=\frac{\mathrm{plug}\mathrm{length}}{\mathrm{speed}} \\ \mathrm{Kinetic}\mathrm{injection}\mathrm{time}\left(\mathrm{t}\right)=\frac{1}{\mathrm{Applied}\mathrm{electric}\mathrm{field}} \end{gathered}$$

To explain the electrokinetic injection time, when decreasing of applied voltage by a factor of 2. The electro kinetic injection time is

$$\begin{gathered} \mathrm{Time}\left(\mathrm{t}\right)=\frac{\mathrm{plug}\mathrm{length}}{\mathrm{speed}} \\ =\frac{\mathrm{plug}\mathrm{length}}{{\mathrm{μ}}_{\mathrm{app}}\left(\mathrm{E}{\displaystyle \frac{{\mathrm{K}}_{\mathrm{h}}}{{\mathrm{K}}_{\mathrm{c}}}}\right)} \end{gathered}$$

In the above equation, the applied electric field is present as a denominator so 2 fractions of applied electric field is doubly increased the electro kinetic injection time.

$\mathrm{Kinetic}\mathrm{injection}\mathrm{time}\left(\mathrm{t}\right)=\frac{1}{\mathrm{Applied}\mathrm{electric}\mathrm{field}}$

On the other hand, the time is inversely proportional to the applied electric field hence, the decreasing of half of the applied electric field is results the doubly increasing kinetic injection time.

Since the kinetic injection time is inversely proportional to applied electric field, injection time is doubled when applied voltage is halved.