Question

please show all work, answers are in red

2. Three point charges [q10 nC, 15 nC, q 5 nC] are placed at the vertices of a 5nC rectangle, as shown. (a) Find the magnitude and direction of the net cm.- electric force exerted on charge q2-15 nC due to the two other charges. 3cm (b) Determine the magnitude and direction of the net electric field due to the three charges at the vacant vertex, as shown (c) What electric force will be exerted on a fourth10nC 15 nC charge, q4-15 nC, placed at the same vacant vertex? I(a)11.3 x 10 N at 222, (b) 9.8 x 10 N/C at 224°, (c) 1.47 x 10 'N at 44]

Answer 1

(a) F_21 = 4 Vertical-bar q_2 Vertical-barVertical-bar q_1 Vertical-bar /0.04^2 = (9 times 10^9) (15 times 10^-9) (10 times 10^-9)/0.04^2 = 8.4375 times 10^-9 N F_23 = 4 Vertical-bar q_2 Vertical-barVertical-bar q_3 Vertical-bar /0.03^2 = (9 times 10^9) (15 times 10^-9) (5 times 10^-9)/0.03^2 = 7.5 times 10^-9 N Therefore Vertical-bar F_2 Vertical-bar = squareroot F_21^2 + F_23^2 = 1.13 times 10^-3 N or 11.3 times 10^-4 N theta = 180 + tan^-1 (F_23/F_21) = 1222 degrees (b) d = squareroot 0.03^2 + 0.04^2 = 5 cm = 0.05 m theta = tan^-1 (3/4) = 36.9 degrees E_1 = (9 times 10^9) (10 times 10^-9)/0.03^2 = 100,000 N/C E_2 = (9 times 10^9) (15 times 10^-9)/0.05^2 = 54,000 N/C E_3 = (9 times 10^9) (5 times 10^-9)/0.04^2 = 28,125 N/CVertical-bar E_x Vertical-bar = E_3 + E_2 cos theta = 71,308 N/CVertical-bar E_y Vertical-bar = E_1 - E_2 sin theta = 67,577 N/CVertical-bar E Vertical-bar = squareroot E_x^2 + E_y^2 = 9.8 times 10^4 N/C Direction = 180 + tan^-1 (E_y/E_x) = 224 degrees