P22_005

# P22_005 - 2 q on +2 q is ~ F 3 = k (2 q )(2 q ) a 2 i . (a)...

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5. We put the origin of a coordinate system at the lower left corner of the square and take + x rightward and + y upward. The force exerted by the charge + q on the charge +2 q is ~ F 1 = k q (2 q ) a 2 ( ˆ j) . The force exerted by the charge q on the +2 q charge is directed along the diagonal of the square and has magnitude F 2 = k q (2 q ) ( a 2) 2 which becomes, upon Fnding its components (and using the fact that cos 45 =1 / 2), ~ F 2 = k q (2 q ) 2 2 a 2 ˆ i+ k q (2 q ) 2 2 a 2 ˆ j . ±inally, the force exerted by the charge
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Unformatted text preview: 2 q on +2 q is ~ F 3 = k (2 q )(2 q ) a 2 i . (a) Therefore, the horizontal component of the resultant force on +2 q is F x = F 1 x + F 2 x + F 3 x = k q 2 a 2 1 2 + 4 = ( 8 . 99 10 9 ) ( 1 . 10 7 ) 2 . 050 2 1 2 + 4 = 0 . 17 N . (b) The vertical component of the net force is F y = F 1 y + F 2 y + F 3 y = k q 2 a 2 2 + 1 2 = . 046 N ....
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