# Magnitude of charge

1. Jun 5, 2007

### bidhati

1. The problem statement, all variables and given/known data

Two charges Q1 and Q2 of equal magnitude but opposite signs are fixed a distance 5.00 m apart in a vacuum The line PQ is a 12.0 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 = 1.00 × 10−2 s the electron is observed to pass point Q. Determine the magnitude and sign of both Q1 and Q2

2. Relevant equations

Fel = 1/ 4 pi E q1q2 / r^2
a = 2s / t^2

3. The attempt at a solution
well basically I think I need another equation to combine with coulomb's law so I can make Q the subject and thus find the magnitude of charge.

I know that Fel is inversely proportional to r^2 and that if I find Q I just need to halve the value as both charges are equal.
as for the last part of the question electron = - so Q2 = + and thus Q1 = -

obviously there is something I am missing here a helpful pointer please?

2. Jun 5, 2007

### Amith2006

I think u need find the change in electrostatic P.E as it moves from P to Qwhich will give an equation in q. Equate this to the kinetic energy gained. Hope this helps.

3. Jun 5, 2007

### chaoseverlasting

Well, the electric field changes with time... as the charges are of opposite signs, they will accelerate towards each other and hence the magnitude of the field will change... You need to find the change in the field and then calculate the variable force on the electron.

4. Jun 6, 2007

### Amith2006

The electric field is uniform along the length PQ.

5. Jun 6, 2007

### f(x)

Since the charges are opposite, the +ve carge Q1(lets say) will be towards Q and the negative charge will be towards P. Now calculate the total force acting on the electron , repulsion due to Q2 and attraction due to Q1. Thus you'll get the acceleration of the electron. This will help solve for the unknowns