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Quick Answer: 

$10.1\dfrac{kQ^2}{l^2}, 61^\circ \textrm{ above the positive } x \textrm{ axis.}$

Giancoli 7th Edition, Chapter 16, Problem 15

(7:17)

Chapter 16, Problem 15 is solved.

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Comments

Why are we looking for d^2 when finding the diagonal distance? The distance of the hypotenuse of a 3,4,5 triangle is not 25 its square root of it which is 5..... A bit confused:(

Hi nategrana, thanks for the question. To determine the 'x' component of the force on the '2Q' charge, we need the 'x' component of the force due to the '4Q' charge diagonally opposite. The force formula has the distance between the '2Q' and '4Q' charge in the denominator, but it is squared. We're looking for the square of the distance since that's the quantity to plug into the force formula.

Hope that helps,
Mr. Dychko

WOA! Totally missed that. Thank you Mr. Dychko. Great site!

Thanks! Good luck with your studies!