Question

Please answer all questions and show all work and reasoning. Thank you!! :)

2. A spring is attached to a object with charge q +3.00 uC on a frictionless horizontal surface. The spring has some equilibrium length L. A charge q2 -9.00 HC is placed 4.00 cm away from the positive charge. In response, the spring stretches and reduces the distance between charges to 3.70 cm. Find the spring constant k. a. 1 1 72 b. If a charge q +5.00 uC had been placed in the same position, what would the spring's displacement have been? Include direction.

Answer 1

Q_1 = 3 times 10^-6 C q_2 = -9 times 10^- 6 C d = 0.04 m d_1 = 0.037 m (a) F.B.D of q_1 F_k = Fe (k) (d - d_1) = (k) ( Vertical-bar q_1 Vertical-bar ) ( Vertical-bar q_2 Vertical-bar )/d^2_1 K = (9 times 10^9 (3 times 9) (10^- 12)/(0.037)^2 (0.04 - 0.037) K = 59167 N/M (b) d = 0.04 m q_1 = 3 times 1\1)^- 6 c q_2 = 5 times 10^- 6 c F.B.D of q_1 F_k = Fe \r\n (K) (d_1 - d) = (k) (q_1) (q_2)/d^2_1 (d_1 - d) (spring�s displacement) = (9 times 10^9) (3 times 5) (10^- 12)/d^2_1 (59167) d^3_1 - (0.04) d^2_1 - 2.3 times 10^- 6 = 0 d_1 = 0.0400574 m spring�s displacement = (d_1 - d) m = 0.0000574 m = 5.74 times 10^- 5 m