(+Q)-- -- -- -- (-Q) : : : : (-Q)-- -- -- -- (+Q) four point charges, each of th

Question

(+Q)-- --   --   -- (-Q)

     :                   :

     :                   :

(-Q)--   --   -- -- (+Q)

four point charges, each of the same magnitude, with varying signs are arranged at the corners of a square of as shown. In what direction will the net force that acts on the charge at the upper right corner be directed, this is the (-Q).

I know that the net force will be pointed diagnally inward towards the other -Q and you can use vector addition and coulombs law but I am not sure how to show it mathmatically to prove this.

Sorry if this is confusing, any help is greatly appreciated.

Explanation / Answer

So you need to do a free body diagram on the top right charge and then add the vector sum of the forces. A ............ B . . . . C.............D FD (force from D) = k Q^2/d^2 pointing downwards (south) FC = k Q^2/ 2d^2 pointed 45 degrees north of east. FA = k Q^2/d^2 pointed to the left (west) now you need to get the two components Y: kQ^2/d^2 - sqrt(2)kQ^2/4d^2 = (4-sqrt(2)) kQ^2/d^2 pointed downwards x: kQ^2/d^2 - sqrt(2)kQ^2/4d^2 = (4-sqrt(2)) kQ^2/d^2 pointed to the west now take the inverse tangent of y/x to get the angle (you know that the force points south west based on the components) Since they are the same magnitude, the tangent of that angle will be 45 degrees. so it does point diagonally inwards at 45 degrees south of west.

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