Electric Forcs and Work (Algebra based physics)

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SUMMARY

The discussion focuses on calculating the work required to move two electric charges, q1=+7.15 x 10-5 C and q2=-7.80 x 10-4 C, from a distance of 71.5 cm to 31.0 cm apart. The user initially applied the formula F = kq/r but miscalculated the effective distance (r) as 0.513 m. The correct approach involves understanding electric potential energy and the work-energy principle. The user ultimately calculated the force as 968 J and the work done as 300 J, indicating a misunderstanding in the application of these concepts.

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  • Familiarity with the work-energy principle in physics
  • Basic algebra skills for manipulating equations
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Seikon85
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Charges q1=+7.15 x 10-05 C and q2=-7.80 x 10-04 C are placed at rest 71.5 cm apart. How much work must be done by an outside agent to move these charges slowly and steadily until they are 31.0 cm apart?

Ok so I tried a few things on my own, and got stuck. So I referred to a previous post from this morning for guidance:

http://answerboard.cramster.com/physics-topic-5-199660-0.aspx

however, I am still getting the wrong answer.


I am using F = kq/r initially. The r value i am using was obtained by figuring out the distance one charge moved. (.715-.31)/2. Then subtracting that distance from .715. So my r = .513m here.

I get F = 968J.

W = F*d = 968J * (.31m) = 300J.


Am I missing something obvious. Any insight?

Thanks.
 
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