Question

Three charges (91 = 6.1 C, -3.9 C, and q3 1.1 C) are located at the vertices of an equilateral triangle with side d 6.1 cm as shown in Figure 1 42 d Figure 1 (a) What is the r-component of the net force on charge q3? (3pt) (b) What is the y-component of the net force on qs? (3pt) now added as shown in Figure 2. What is the r-component (c) A fourth charge q4 1.1 C is of the new net force on 2? (4pt) d 2 d d (94) Figure 2: (d) What is the y-component of the new net force on q2? (4pt) пеw (e) What is the r-component of the new net force on (f) How would you change qi (while keeping q2, s and q fixed) in order to make the net force on q2 equal to zero? (2pt) n? (4pt) (A) Increase its magnitude and keep its sign the same (B) Decrease its magnitude and keep its sign the same (C) Increase its magnitude and change its sign (D) Decrease its magnitude and change its sign (E) There is no change you can make to qi that will result in the net force on q2 being

Solve and Explain please!

Answer 1

Given q = 6.14C q = -3-9 uc Q = l l Mc da 6.1cm Ган да Ro Fon Faire 8324 6o ar y Fair F31 - Force on 3 due to I F32 - force on 3 due to a Fai = KG 34 - 9x10x1x106x6:1810-6 (6.1810-29 2 For = 16.229 N 12s F32 = k L32 - 9X100X11X106x3.98106 (6.1810-292 F32 = 10.376 N

@ I - X-Component net force on hs is Fru F₂ cosa + Fzz Cos Q. Fax = 16.22 98 605 66 + 10.346 x cos 60° Fax = 13.302 N. y - Component of net Zo is Fzy = F3, sing - Faasing- Fay = 16:229 x sin30 – 100376 xsin 30° @ F3y = 5.068 N T 9= H UC is introduced Now the F23 23 234 Faun o coa [ 521 = 88 54.54 N Fan Go VA + F293 = 9x10x 3.9 X 11X1Ot12 H = 10.376 (6.1X10-22

Therefore, un component of new net & on Iz is, Frau = Fait Fazut Faun = 57.54 + F23 los 600 t Fay CosGo Fax = 57.547 10.376 Fru = 67.9IN -@y-component of new net force on qiz F2y = Fazy t-Fany) Fay = Fes sin 66 +(Fey sin 600) @ on a To calculate new net force We found 600 A a DA . Fiz = K2, 22 t Fiy = 149 In = kain 22 - Kh24

Fax = 25:08 N D is the correct answer.