Question

KHwork 01 Electric Force and Field Problem 1 Constants Learning Goal: To practice Problem-Solving Strategy 21.1 Coulomb's Law Three charged particles are placed at each of three comers of an equilateral triangle whose sides are of length 2.6 cm. Two of the particles have a negative charge: q1 -6.7 nC and g2-13.4 nC The remaining particle has a positive charge, 93 8.0 nC. What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Answer 1

Given and side of triangle = 2.6 cm r = 2.6 times 10^-2 m We know that k(coulombs constant) = 9 times 10^9 Nm^2/c^2 F_3, Force acting on q_3 and q_1 = kq_1 q_2/r^2 F_31 = 9 times 10^9 times (-6.7) times 10^-9 times (-13.4)10^-9/(2.6 times 10^-2)^2 = 808.02 times 10^-9/6.76 times 10^-4 = 119.52 times 10^-5 N Force on due to q_2 F_32 = k q_2 q_3/r^2 = 9 times 10^9 times (-13.4) times 10^-9 times 8 times 10^-9/(2.6 times 10^-2)^2 = -964.8 times 10^-9/6.76 times 10^-4 = -142.72 times 10^-5 N \r\nW.K.T F = F_x + F_y F_x = F_32 along x-axis (single Force) But F_y rightarrow F_31 is broke into = parts (F_31 cos theta(along x) & F_31 sin theta (along y-axis)) At x component F_31 = F_31 cos theta In an equilateral triangle theta = 60 degrees F_31 = 11952 times 10^-5 cos 60 degrees = 59.76 times 10^-5 N at y component