Question

Your answer is partially correct. Try again. Two point charges are lying on the y axis as in the figure: g- 320 pC and q +320, С they are equ distant from the pont e aneh les on the r axis s nans the magnitude of net electric field at P? (b) A small object of charge do 7.60 uC and mass m 1.27 g is placed at P. When it is released, what is the magnitude of its acceleration? 91 0.630mm 322 0.630 m 42 (a) N (b) Num AW Rights Reserved A Division of

Answer 1

a) q_1 = -3.2 mu c q_2 = 3.2 mu c theta = 32.2 degrees d = 0.630 m The net electric field at a point p is E = E_1 + E_2 E = kq_1/d^2 times [-cos theta i + sin theta j] + kq_2/d^2 times [cos theta i + sin theta j] E = 2(kq/d^2) sin theta j E = 2 times 9 times 10^9 N.m^2/c^2 times 3.2 times 10^-6 c/(0.630 m)^2 times in 32.2 degrees j E = 7.733 times 10^4 N/C j magnitude of E = 7.733 times 10^4 N/C b) q_0 = 7.60 mu c = 7.60 times 10^-6 C m = 1.27 g = 1.27 times 10^-3 Kg F = q_0 E and F = ma implies a = F/m acceleration a = q_0 E/m = 7.60 times 10^-6 C times 7.733 times 10^4 N/C/1.27 times 10^-3 kg a = 462.783 m/s^2 m = 0.012 kg, q = -16 mu c