Question

Two charges, qı = 40 nC and q2 = -10 nC are instantaneously located on the x-axis a distance x = 0.10 m apart as shown in the figure below. The two charges are moving with velocity v = 1.30 Mm/s in the +z direction. There is nothing at the point P a distance y = 0.70 m above the x-axis in the +y direction. Y A. 2 -92 Х NI

Answer 1

2 q q2 V we have 26= 0 lm 4X10-bc. g=0.7m -10-?c ca? 07 ds> - E GITE 3 LAE P due to q2, folo 7m. ña E = -9x10²x10 (0.7202 3)Ę=-183.67 Nici due to q. I a Colêt 0.7 ŷ, Electric field de to To a to unfo (0.71)3 f (Mila No. 2-40.7) = 0.71 So y component of Е, = 1, ° Ч Хto-® ҳо:7 = 712.769 N (co So Resultant at P in ŷ direction = E + Ey (712-7658367) = ($29.03.) Somat Pe y comp of Electre fille = 529-0.3. N{c.

(B) Magnetic field due to point change moving with velocity v given by : Mo q uxa YT with velocity (8) Magnetic field from Bist-Savaut law due to point change of moving given " by as Mo q ux So, for he is u and h. y vendimentim y mequite field (@XE) 2-5 do y component magnetic field due to a for q. % = 0.1sê toify VO:1) +0.7) Since y = v component will be given only by Câ x û so (dB) y a to 21 (12(0:1) Foto:72 X+4x100-2X 1-3 610-3x0.1 7 So 8 3/2 2 ks -- -7 To 0.35 .18 1.47 X 16 Ti it So 143 X 10' at gp gey y component of materia di hotel O Magnetic field at Plain directio (xa). ( x 9) = - - so second optim -; qorut.