Posted by leess on .
a +35x10^6 C point charge is placed 32 cm from an identical +32x10^6 C charge. how much work would be required to move a +50.0x10^6 C test charge from a point midway between them to a point 12 cm closer to either of the charges?
please show work and explain. i still don't understand what Damon said. we didn't learn how to solve it with integrals. is there another way
physics  Damon, Friday, February 22, 2008 at 4:03pm
q is each of our two charges
Q is our test charge
Left charge at x = 0
Right charge at x = .32 m
force due to left charge = k q Q/x^2
force due to right charge = k q Q/(.32x)^2
when x = .16, the middle, the sum of those two forces is zero. However as you move off center, the one nearer will push back harder. We move to x = .16+.12 = .28. At that point we are .32 .28 = .04 from the right charge.
The integral of dr/r^2 = 1/r =[1/Rend 1/Rbegin] = [1/Rbegin  1/Rend]
Let's assume we move right (+x direction)
Work done against left charge (negative because it is pushing in the direction of motion so we are holding back) =  k q Q( 1/.16 1/.28)
Work done against right charge = +k q Q(1/.04 1/.16)

physics 
Damon,
If you did not have the integrals to derive the equation, I bet you were given an equation like this:
Potential Energy of charge close to other charge = PE or U = k Q1 Q2 / R
That is all I did with the integrals.
That is where the 1/Rend and 1/Rbegin comes from.
The work done is just the change in potential energy.
Since I knew how to integrate, I did the force times distance, but I bet the result of my integral, the 1/R equation, is in your book. 
physics 
joe,
1.3