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April 1, 2015

April 1, 2015

Posted by **leess** on Saturday, February 23, 2008 at 2:42pm.

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)

- physics -
**Damon**, Saturday, February 23, 2008 at 5:08pmIf 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**, Wednesday, July 7, 2010 at 11:01pm1.3

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