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haganmc
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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?
 one year ago
 one year ago
haganmc Group Title
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?
 one year ago
 one year ago

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haganmc Group TitleBest ResponseYou've already chosen the best response.0
dw:1346286892631:dw
 one year ago

bhaweshwebmaster Group TitleBest ResponseYou've already chosen the best response.1
dw:1346298352562:dw
 one year ago

haganmc Group TitleBest ResponseYou've already chosen the best response.0
the top right is also positive
 one year ago

haganmc Group TitleBest ResponseYou've already chosen the best response.0
dw:1346298573822:dw
 one year ago

haganmc Group TitleBest ResponseYou've already chosen the best response.0
dw:1346298617109:dw
 one year ago

bhaweshwebmaster Group TitleBest ResponseYou've already chosen the best response.1
dw:1346299022875:dw This is the real figure.. and the resultant force is along the diagonal but repulsive..
 one year ago

haganmc Group TitleBest ResponseYou've already chosen the best response.0
the answer is \[(2\sqrt{2}) \frac{ KQq }{ L ^{2} }\].. i'm still confused on how to get it
 one year ago

bhaweshwebmaster Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

haganmc Group TitleBest ResponseYou've already chosen the best response.0
alright thanks!
 one year ago

amallari Group TitleBest ResponseYou've already chosen the best response.0
Hi, did you have to draw vectors for this question? I can't seem to get it right.
 one year ago
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