Question

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 q,--9.00 μC 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. 2. a. Find the spring constant k. 91 92 If a charge q2 = +5.00 μc had been placed in the same position, what would the spring's displacement have been? Include direction. b.

Answer 1

q_1 = 3 times 10^-6 C q_2 = -9 times 10^-6 C d = 0.04m d_1 = 0.037m (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 10^2) (3 times 9) (10^-12)/(0.037)^2 (0.04 - 0.037) K = 59167 M/m (b) d = 0.04m q_1 = 3 times 10^-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 springs displacement = (d_1 - d)m = 0.0000574m = 5.74 times 10^-5 m