com www.GOALias.blogspot.com

www.GOALias.blogspot.comElectric Chargesand FieldsThus,3 QqForce F 1on Q due to charge q at A = along AO24πε0l3 QqForce F 2on Q due to charge q at B = 24πεalong BO0 l3 QqForce F 3on Q due to charge q at C = 24πεalong CO0 l3 QqThe resultant of forces F 2and F 3is 24πεalong OA, by the0 l3 Qqparallelogram law. Therefore, the total force on Q = 2( rˆ− rˆ)4πε0l= 0, where ˆr is the unit vector along OA.It is clear also by symmetry that the three forces will sum to zero.Suppose that the resultant force was non-zero but in some direction.Consider what would happen if the system was rotated through 60ºabout O.EXAMPLE 1.6Example 1.7 Consider the charges q, q, and –q placed at the verticesof an equilateral triangle, as shown in Fig. 1.10. What is the force oneach charge?FIGURE 1.10Solution The forces acting on charge q at A due to charges q at Band –q at C are F 12along BA and F 13along AC respectively, as shownin Fig. 1.10. By the parallelogram law, the total force F 1on the chargeq at A is given byF 1= F ˆr 1 where ˆr1 is a unit vector along BC.The force of attraction or repulsion for each pair of charges has the2qsame magnitude F =24 π ε0lThe total force F 2on charge q at B is thus F 2= F ˆr 2, where ˆr 2is aunit vector along AC.EXAMPLE 1.717