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November 23, 2014

Homework Help: physics

Posted by leesa on Saturday, February 23, 2008 at 12:23pm.

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Homework Help Forum: physics
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Posted by leesa on Friday, February 22, 2008 at 3:32pm.

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)

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