Question

Due: Online via Canvas on Tue., Jan. 15 before 11 a.m. Two small, identically charged, Styrofoam spheres are hung from the ceiling using strings, and reach equilibrium as shown in Figure 1, below. Each sphere has a mass of 0.0300kg and an unknown positive charge +0, so they repel each other due to the electric force. The tension in the string is an unknown, 1. length of each string is 0.150m. The angle of each string, relative to vertical, is 0 4.50°. One way to calculate the unknown charge is via the steps of a free-body diagram analysi work as you work through the steps below. Significant figures will only be graded for the final answer for the unknown charge, so you may keep a couple of extra digits in the intermediate steps. s. Show all your First, calculate the distance between the spheres. Draw a free-body diagram for one of the spheres. Make sure to label all the forces present. Write a symbolic expression for the sum of the x-components of all the forces on that sphere (Fot Then, using Newton's 2 nd Law Fet-ma, calculate and sub in the value of Fnet Write a symbolic expression for the sum of the y-components of all the forces on the sphere (Fnet.y). Then, using Newton's 2nd Law Fnety-may, calculate and sub in the value of Fnet. Algebraically solve the two equations from c) and d) by elimination or substitution in order to deduce the magnitude of the electric force on that sphere (it is optional to also solve the magnitude of the tension in your x, y equations). Now that you know the magnitude of the electric force, treat it as a given, and apply the formula for the electric force to numerically calculate the unknown charge, +Q. y. . +Q Figure 1

Answer 1

L = 0.150 cm m = 0.030 kg theta = 4.50 degree (a) DISTANCE between two SYNERES x = 2 AC = 2 L sin theta x = 2 (0.150) times ln (4.50 degree) x = 0.0235 m (c) F_net, x = F_e - T sin theta m a_x = F_e - T sin theta a_x = 1/ m (F_e - T sin theta) (d) F_net y = mg - T cos theta a_y = 1/ m (mg - T cos theta) (e) a_x = a_y = 0 F_e = T sin theta, F_e = 0.295 sin theta 4.50 degree F_e = 0.023 N T cos theta = mg T = mg/ cos theta = 0.0300 times 9.8/ cos 4.50 degree T = 0.295 N (d) F_e = 1/ 4 pi E_o theta^2/ x^2 implies theta = [(0.023) times 0.0235/ 4 times 10^9]^1/ 2 theta = 3.76 times 10^- 8 C therefore 1/ 4 pi = 8

I will be glad to see your comment if you have any query and thumb up if you are satisfied. Thanks...