Question

A 13.813.8 kV/m electric field is used to inject a sample onto a 52.852.8 cm long capillary that has a diameter of 50 μm. The sample has an apparent mobility of 2.57×10−8 m2/(V⋅s)2.57×10−8 m2/(V⋅s) and a conductivity that is 1/5 of the conductivity of the buffer. How long must the electric field be applied to inject a sample that is 3.263.26% of the length of the capillary?

t=

Answer 1

The electrophoretic mobility $\mu _{p}$ of the analyte or sample can be expressed as a function of migration time and the field strength:

$\mu _{p} = \frac{L.L_{t}}{t_{r}.V} = 2.57*10^{-8} \frac{m^{2}}{V.s}$

L : is the distance from the inlet to the detection point = 3.263 % of L_t

L_t : is the total length of the capillary = 52.8528 cm = 0.5285 m

t_r : is the time required for the analyte to reach the detection point (migration time)

V : is the applied voltage (field strength)

= E(ElectricField) = 13.813- m

Substituting the respective values to find t_r

8 m2 0.0326 * 0.5285(m) 2.57 * 10 Ts=t(s) * 13813)

Solving for t_r :

= 48.533 seconds

Hence the Electric field must be applied for 48.533 sec to inject a sample that is 3.263% of length of capillary