Resolução - Halliday - Volume 3 - Eletricidade E Magnetismo - P22 005

Transcript

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 +2q is q(2q) F1 = k 2 (−ˆj ) . a The force exerted by the charge −q on the +2q charge is directed along the diagonal of the square and has magnitude q(2q) F2 = k √ (a 2)2 √ which becomes, upon finding its components (and using the fact that cos 45◦ = 1/ 2), q(2q) ˆ q(2q) ˆ i+k √ j . F2 = k √ 2 2 2a 2 2 a2 Finally, the force exerted by the charge −2q on +2q is (2q)(2q) ˆ F3 = k i . a2 (a) Therefore, the horizontal component of the resultant force on +2q is 
q2 1 Fx = F1x + F2x + F3x = k 2 √ + 4 a 2    −7 2
  1.0 × 10 1 √ = 8.99 × 109 + 4 = 0.17 N . 0.0502 2 (b) The vertical component of the net force is Fy = F1y + F2y + F3y = k q2 a2
1 −2 + √ 2  = −0.046 N .