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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?
 2 years ago
 2 years 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?
 2 years ago
 2 years ago

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

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

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

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

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

haganmc Group TitleBest ResponseYou've already chosen the best response.0
alright thanks!
 2 years 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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