Physics
posted by sandhu on .
Two point charges q1 and q2 are held 4.00 cm apart vertically. An electron released at the middle point that is equidistant from both charges undergoes an initial acceleration of 8.95 X 10^18 m/s2 directly upward,parallel to the line connecting q1 and q2 .Find the magnitude and direction of q1 and q2

Two point charges q1 and q2 are held 4.00 cm apart vertically. An electron released 3 cm from the middle point that is equidistant from both charges undergoes an initial acceleration of 8.95 X 10^18 m/s2 directly upward,parallel to the line connecting q1 and q2 .Find the magnitude and direction of q1 and q2

It is not certain to me that the electron "middle point" is on the line connnecting the two charges, or not.
If it is not on that line, then the acceleration upward tells us that the horizontal components of the force on the charge are equal and opposite, that is no resultant force horizntally.
If they are equal and opposite horizontally, they must be equal forces vertically, that is, the lower charge is , and the upper +. You cant do any more without knowing distance, as I see it.
Now if the electron is on the line connecting the charges, you can solve it, as Etotal is due to equal parts from each charge, and you know the distance (.02m) 
Ok, I see you added the horzontal distance. From that, you can find the distance to each charge (PYTH theorm) and then find E (remember both charges contribue 1/2 E).