Question

18.48 - Conductors and Electric Fields in Static Equilibrium Puntos:2 Find the total Coulomb force on the charge q in the figure below, given that q = 1.00 μС, qa-1.90 μC, qb =-2.90 μС, qc--4.30 με, and qd = 1.00 μC. The square is 60.0 cm on a side Ga go Magnitude of force -0.0707N Enviar Respuesta Incorrecto. Tries 2/7 Intentos Anteriores Angle of force (wrt +x-axis) Enviar Respuesta Tries 0/7

Answer 1

F_net = k/r^2 [[ Vertical-bar q q_a Vertical-bar c is 315 degree] + [ Vertical-bar q q_b Vertical-bar c is 45 degree] + [ Vertical-bar q q_c Vertical-bar c is 225 degree] + [ Vertical-bar q q_d Vertical-bar c is 135 degree]] r = diagonal of square/2 = squareroot 60^2 + 60^2/2 times 10^-2 m r = 60 squareroot 2/2 times 10^-2 = 30 squareroot 2 times 10^-2 m = 0.3 squareroot 2 m = 0.424 m therefore r = 0.424 m therefore F_net = (9 times 10^9)q/(0.424)^2 [(1.90 mu) [1/squareroot 2 i bar + 1/squareroot 2 j bar] + (2.9 mu) [1/squareroot 2 i bar + 1/squareroot 2 j bar] + (4.3 mu) [-1/squareroot 2 i bar - 1/squareroot 2 j bar] + (1 mu) [-1/squareroot 2 i bar + 1/squareroot 2 j bar]] F_nat bar = (9 times 10^9) (1 times 10^-6) times (10^-6)/(0.424)^2 times squareroot 2 [(1.9) (i bar - j bar) + (2.9) (i bar + j bar) + (4.3) (i bar - j bar) + (1) (-i + j)] F_net bar = 35.399 times 10^-3 [-0.5i bar - 2.3j bar] Vertical-bar F_net bar Vertical-bar = 83.319 times 10^-3 N Vertical-bar F_net bar Vertical-bar = 0.083319 N Angle theta = tan^-1(-2.3/-0.5) = 77.7 degree