Question

Part A, B, and C

<Chapter 17 Problem 17.66 ③ got to Review Part A Acharge of -2.80 nC is placed at the origin of an ty-coordinate system, and a charge of 2.35 nC is placed on the yaxis at y = 3.75 cm. If a third charge, of 5.00 nC, is now placed at the point 2.90 cm.3.75 cm find the candy components of the total force exorted on this charge by the other two charges. Enter your answers in newtons separated by a comma. VALD O F. Fy- Submit Request Answer Part 8 Find the magnitude of this force Express your answer in newtons

Answer 1

9,- -2.50nc :- 2.80 x109 9,- 9.35n0 = 2,35x109 9190CM 9. = 5.000 5.0 X10 4,0 ?.. fo with F 79= 3.75 cm = 0.0375m 0375m 3.750 7g3 = 9.90cm = 0.029 m . >83 = 5 mx + r y figure o Yig = 120.8375)*7(0.0937.2 0.0474 m Part @ . force exerted on the charge by other charge will be F= k19,2.1 .; Kette = 9x09 mailem 41 - a fore exerted on az by q, Fi 3 k19,831 813 9x109 x 1(-280x107) (5.0x109) & 10.0474) F, 5.61x165 N ® Force exerted on az by 92 for k1qa231 9x108x 2.35X109 x 5x1071 az (0.029) fg = 19.57 xron -

a> from Ligure , x,y directions for fi = Fix ? Fiy -for fiq es fa it is only in the x-- direction all the forces x,y directions → So let us combine seperated by Fx = (- Fx + F2x)} - fy = (– Fry) i = 0 => Fix= of cowo – = 0.6118 0.0375 -0.79114 Fly = F, sino - 0.029 (823) - from figure cose = 0.0474 (Vxs) sino = = 0.0474 How > fx = ( - F, cose + Fe] +1-(5.61x105 x 0.6118) +(12.578799|| Fe = 9.138xás ne] .. > Fy = (- fy) =(-a sino] af 5.61X105X0.79114] [ty = -4.44X105 ]

Pas fig = .13620 *o, - 4.44 705) 9.138 X - 4.44 710 ? The magnitude of force is F = Jatty = 1 19.138x16057% (-4.44X105)? .IF = 10.16x16sln Part 6 $ . for direction tanda 3 tem o = -4.44x07s 9.138 xiots * = -25.99 ->$= 25.9°) below the *x-arts low the ti-axis