Question

Question 1 2.5 pts Three charges Q1 =-14.0nC, Q2 = Q3 = +4.19C are placed on the vertices of a square of side 2cm as shown in the figure below. S 2cm Find the magnitude of the electric field at the empty vertex (Do not round up intermediate values. Only round your final answer to the nearest whole number

Answer 1

as 0:0,20m 16.02118M- Berity on is a Let I electric field due to charge 2, , I, and are E, a Eg and Ez respecterly. 2,= -14 nC 2 = 2 = +401 nC Side of square = 2 cm = 0.029 m from pythagosius throom BD = J (AB) + (ABP = 110.2)? +(02)2 => BD = (0.0 M2 m

Magnitude of electric field due to charge a at point d' E = k19,1 I (BD)? 19,1 = lync=14810-9c BD= 0.02 x rã m - 2 k= 9x109 x 10-12 m => E, = 9x 109 x 14 XIOM 186 = (ora x108) 3 2 8 x10-4 => Ei= 15. 75 x1074 NC

for charge 2 1921= 4oncs 4.1X10-90 CD = 2 x10m -> Es = 9x69 x 4.1 xlor"36.9 (2x10-22]? 4x10-4 => Ez = 9.225 XION/ since, 23 =2 and AD= CD =) Ez = k1831 (MD)Z = 611.) – to TCD)2 = Ep => Ez = 9.225 x16 NC

- so electric fields at point de WV Let Eaz is resultant of Ez and Ez electric field. Since E, and Es are equal in magnitude and are perpendicular to each other, angle between Egg and Es is o and tan O = E2 = 1 [Ea - Ez] => 0 = 450 E3 so electric fields E, and Eqz are colinear. so magnitude of total electric field at D is E 1 E - Egal

1 sum vector sum of Ez and Es 1 = Ea3 = 5 6 + 6 = 12 tahun [E3 = Ez] → E23 = 12 Ey = x 9.285 x10" NIC = 13.05 X10CNIC from 0 total electric field - Exotal = 1 E,- Egol Etotal = 15,75 x 10"– 13.05 x 10" - Etotal = 2.7038 N/C total - Exotal = 3 N/C [Nearest Whole Number ]