Electric Forces Two charges of equal magnitude but opposite signs?

Question

Two charges Q1 and Q2 of equal magnitude but opposite signs are fixed a distance 4.00 m apart in a vacuum The line PQ is a 10.0 cm section of the line joining the two charges and is placed centrally between them. Over the distance PQ the electric field may be taken to be uniform. An electron is released, with negligible initial speed, from point P at time t = 0. At t = 1.50 × 10−2 s the electron is observed to pass point Q. Determine the magnitude and sign of both Q1 and Q2.m

Answer 1

By supper position
Et=E1+E2
Et=k(Q1 + Q2)/R^2
k - constant
Q=Q1=Q2 charges
R - distance half way between the charges

Et=2 kQ/R^2

Et=qF
F=ma
a=2S/t^2

Et= 2q mS/t^2 = 2 kQ/R^2

Q= q m S R^2 / (k t^2) Where

q= 1.60 E-19 C
m=9.11 E–31 kg
S= .1m
R= 2 m

k = 8.98 E+9 N m^2 /C^2
t = 1.50 E−2 s

Q1=Q2 and
The charge to which electron is moving just have to be positive.