Question

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/(.32-x)^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)

Answer 1

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.

Answer 2

1.3