Unformatted text preview: q  ) ( Q ) a 2 cos 45 â—¦ Â¶ = 1 4 Ï€Îµ Âµ Q 2 2 a 2 âˆ’ 2  q  Â· Q a 2 1 âˆš 2 Â¶ which (upon requiring F x = 0) leads to  q  = Q/ 2 âˆš 2 or q = âˆ’ Q 2 âˆš 2 . (b) The net force on q , examined along the y axis is F y = 1 4 Ï€Îµ Âµ q 2 (2 d ) 2 âˆ’ 2 (  q  ) ( Q ) a 2 sin 45 â—¦ Â¶ = 1 4 Ï€Îµ Âµ q 2 2 a 2 âˆ’ 2  q  Â· Q a 2 1 âˆš 2 Â¶ which (if we demand F y = 0) leads to q = âˆ’ 2 Q âˆš 2 which is inconsistent with the result of part (a). Thus, we are unable to construct an equilibrium conÂ±guration with this geometry, where the only forces acting are given by Eq. 221....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Force

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