Question

Two charges, q1 = 90 nC and q2 = -10 nC are instantaneously located on the x-axis a distance x = 0.30 m apart as shown in the figure below. The two charges are moving with velocity v = 1.80 Mm/s in the +z direction. There is nothing at the point P a distance y = 0.60 m above the x-axis in the +y direction. A ----) X % ma NI Part A Find the y-component of the net electric field due to the two charges, at the point P. μΑ ? |]= Value Units

Answer 1

Solution : the plane where charge q, inat origin and Plot ry 9 at x = 0.30m E, sino -E, COMO VE2 182=0.60 Given 9. 10-63-43 0.3 given Electric field at point e due to a 9. E, Ix 406 E, - Lx 90x109 2 1143 110 3++(06)2 5 8.98X70 x 90 8109 10.4572 E,= 1798-0 N/e [ Disected along. 9, P shown in figure Elechic field due at P due to a charge 92 Ea * 406 9 -9 = 8.98X10 X1OXID P9, E = 299.4 / 1 shown in figure

Nowo resolve E, in x by direction Esino Ei Eine - El Coso Eix = 1796.0 cos 63.430 Elx = 803.33N/C → Ę Caso Le { 0-ton (2) Eig = E, sind - 181496.0 Sin 63.48° Ely = 1606.32 N/ Net electric field in a direction at point p Ex = Eix Ex=803.33N/C Net electric field in y-direction at point a Ey = E, sino - E2 1606-32 - 249.4 Ey = 1356.92N/C Resultant eleche field due to 9, 49, charge at point p Ep = VE + Ey √1803.33% + 1356.925 TEp = 1576.871/C Direction from the axir tand - Ey 1356.92 Ex 803-3 o=dant (1.689) 2 Ep 4-59.377

AP y component in only by field B, and (b) Magnetic field at p due to an B = No q 9, x?? 4K B = 10 10 10tx goxlox 171181 Sind (RX - 10*x 90210-9x (108 109 R) x ( 0:34+061) (0.64) - 299.2 X10" [ 1-2 70.3 voľCR xi) + 1870-671008 x)] - 899.2810*x 10°C osuĝ+3.0066)] B = (-323-135ix + 161.560) x 10 B = (323-136 + 161.569) x 10-10 Magnetic field at P due to a J2 xg² 10 *x (-408109 x 6-8 110®3) (0-63 (0.41 -50 x 10 10 (R x i 101060) Be = Ho a प 83 -50 x 10 Bez - 10 sox10 J. equal to = 161.56 1 Testa By (e) Direction at due of Magnetic field at e due to be in ti