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haganmc

  • 3 years ago

the Figure shows four charges at the corners of a square of side L. Assume q and Q are positive. What is the magnitude of the net Force on q?

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  1. haganmc
    • 3 years ago
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    |dw:1346286892631:dw|

  2. bhaweshwebmaster
    • 3 years ago
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    |dw:1346298352562:dw|

  3. haganmc
    • 3 years ago
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    the top right is also positive

  4. haganmc
    • 3 years ago
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    |dw:1346298573822:dw|

  5. haganmc
    • 3 years ago
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    |dw:1346298617109:dw|

  6. bhaweshwebmaster
    • 3 years ago
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    |dw:1346299022875:dw| This is the real figure.. and the resultant force is along the diagonal but repulsive..

  7. haganmc
    • 3 years ago
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    the answer is \[(2-\sqrt{2}) \frac{ KQq }{ L ^{2} }\].. i'm still confused on how to get it

  8. bhaweshwebmaster
    • 3 years ago
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    See that F is the force by 2 -Qs on q... each of which is equal to F=Q∗q /(4Πϵ ∗L^2) and are attractive.. And each of these F act on q at 90 degree.. so their resultant is F ' = F*√2 towards the inside of the square .. along the diagonal.. And, the force on q by 4Q is 4Q∗q/(4Πϵ *2d^2) = 2F.. along the diagonal but outside the square.. So the forces 2F and F√2 are opposite .. Hence the resultant force will be 2F−F√2 = (2-√2) F Now put the value of F=Q∗q /(4Πϵ ∗L^2) to get the actual force..resultant. 4Πϵ can be written into a single constant K.

  9. haganmc
    • 3 years ago
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    alright thanks!

  10. amallari
    • 3 years ago
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    Hi, did you have to draw vectors for this question? I can't seem to get it right.

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