iron stomach
11th June 2007 - 06:24 PM
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
mr_homm
13th June 2007 - 04:51 PM
Assuming E is constant, the force and acceleration on the electron will be constant. Since the electron starts at rest and covers distance d=10cm in time t, its acceleration must satisfy the equation d = 12*at^2. Therefore you can calculate a.
Once you know a, F=ma gives you F. Just look up the mass of an electron.
Once you know F, F=Eq gives you E. Just look up the q of an electron, and remember it'snegative.
Once you know E, E=kQ1r^2 + kQ2/r^2 gives you Q1 and Q2. Since they're equal, the equation becomes E = 2kQ/r^2, where r 2m (the distance from each charge to the small region where the electron moves.
You cannot find the signs with the information that is given, because it is not stated whether P is the end of the line nearer to Q1 or nearer to Q2. You CAN say that the electron will move toward the positive charge, so if you know which way the electron travels, that tells you which of 1 or Q2 is +.
Hope this helps!
--Stuart Andrson