Question

I tried figuring out this problem but i cant seem to get the right answer

SerPSE10 22.A. P. 035. 3.57/7.14 points 13. Previous Answers Three point charges are aligned along the x axis as shown in the figure below. 0.500 m 0.800 m 5.00 nC 3.00 nC -4.00 nC Find the electric field at the following positions. (a) (5.00, 0) E 2.14i N/C (b) (0, 5.00) X N/C Need Help? Read It Submit Answer Practice Another Version -17.18 points SerPSE10 22.A. P.038. 14. +11.0 uC) are located on the corners of a rectangle as shown in the figure below. The dimensions of the rectangle are L 15.1 cm Four identical charged particles (q 60.6 cm and W W (a) Calculate the magnitude of the total electric force exerted on the charge at the lower left corner by the other three charges. N

Answer 1

N607 ty 0.84+s? 13 b) The electric field at 1 Plo, sm) Ē= Es+ Ezt Eu i) Es = due to snc =Kq, i , Ap= 5m Apz 9, = 5x109 Es = 98109 x 5xio”= (189) :) Ez due to 3nc & = K22 at o above - = kez (-coso î + sinoj] Bold da - ५०८Snc oroc 300 Coso = 0.8 √0.82 +52 Sino- 5 Jo.82 +52 1. cp2 се 2 § =60-46632 i +1.03489) Ši ii) Eu - due to -unc C (ep² 0.5² +5²) E = kla3l at d below -x Coso = 0.5 527052 EU = k1231(-cosøê - singoj) sinonime 2 = 9,19%95 ) = -0-14187 (0052452) 15 ; ê= 18 ) - 0.166374710398À – 0.141876-1041873-0.303 î + 142 BP2 9X109 91 -0.5 L 2 52

E = 60-308i +1.423) 14. a) the total force on q at a sk i) F = due to a at B ako ale ce in ce at yfo i) 1 1 (O=tan' W) R = due to q atc (Ac= 11+wz) = 13.99 = kq at o below-se = kq² (- cosoi - sino?). AC2 AC2 is Ę => due to q at D = kq-3 = - kad AD²

W? [²+w? COSO Fx K?? E - kigt 3 - Kop (cosoi + sinoj) Az - keyfine the + [nnt seinen ] ) F= -9X109 x(11X106)?66.6062 + 64 or 153 )i + (0151218160840-15 127°) * = 65-675? - 48.443)N The magnitude of F = 75-6752 + 48-492 -- 48.8 N 1. The magnitude of ² = (48-8N) b) The direction = 0 = 180+ tan'( 48049) = 2630 direction » (2630) S-675)