Question

Two identical point charges (q = +5.80 x 10-6 C) are fixed at opposite corners of a square whose sides have a length of 0.320 m. A test charge (q0 = -7.90 x 10-8 C), with a mass of 6.20 x 10-8 kg, is released from rest at one of the corners of the square. Determine the speed of the test charge when it reaches the center of the square.

Answer 1

The easiest way to do this would be with conservation of energy. Find the potential energy where the charge started, then find the potential energy when the charge is in the middle of the square, then subtract the two. The number you'd have would be how much potential energy was converted to kinetic energy. Then use K=(1/2)mv^2 to find what that speed is.

Here's the potential energy at the beginning:
U = -kqQ/r for each charge paired with the text charge. So
U = -(9E9)(5.80E-6)(-7.9E-8)/(0.320)
Put that in your calculator and you have the potential E from one charge. Everything is the same with the other charge, so the total U is just double of that. Ut = 2U.

Now, here's the potential when the text charge is at the center:
all that is different in the equation is that r changed. So what is the distance from the center of a square to one corner? It would be half the length of the diagonal. The diagonal is 0.320*sqrt(2). So take half of that: D=0.320*sqrt(2)/2. Then the new potential energy is:
U' = -(9E9)(5.80E-6)(-7.9E-8) D
Plug that into your calculator. That also needs to be doubled because of two charges. Now find |Ut - 2U'|. Thats equal to K, the kinetic energy. Now solve:
K=0.5mv^2 for v. You have the mass and we just found K. v is your answer