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awvvu
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Homework Statement
4 charges are arranged in the corners of a square of lengths L as follows:
(-Q) --------- (+q)
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(+4Q) ------- (-Q)
What is the magnitude of the force on the (+q)?
The Attempt at a Solution
My answer doesn't match up with my book's.
x-component of the force is, where K is [itex]\frac{1}{4 \pi \epsilon_0}[/itex]:
[tex]F_x = K q(\frac{- Q}{L^2} + \frac{4Q}{2 L^2}) = K q \frac{Q}{L^2}[/tex]
The x and y components of the force are equal by symmetry:
[tex]\vec{F} = K q \frac{Q}{L^2} (\hat{x} + \hat{y})[/tex]
Therefore the magnitude is:
[tex]F = \sqrt{2} K q \frac{Q}{L^2}[/tex]
However, my textbook says the answer is:
[tex]F = (2 - \sqrt{2}) K q \frac{Q}{L^2}[/tex]
I have no idea where they got the "2" term from. I probably just made a careless mistake somewhere, but I can't see it.
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