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Two charged objects of the same mass (m = 1E[-3]kg) hang from a ceiling. The charge on each object is %3D positive. Each object has the same amount of charge (q1 = q2 = +q). Each object is attached to a massless string of length L = 0.25m. The tops of each string are all attached at the same point on the ceiling. The position of each object is given in the table below. All objects are at rest. We have set the origin to be the position of object 1.

Two charged objects of the same mass (m = 1E[-3]kg) hang from a ceiling. The charge on each object is %3D positive. Each object has the same amount of charge (q1 = q2 = +q). Each object is attached to a massless string of length L = 0.25m. The tops of each string are all attached at the same point on the ceiling. The position of each object is given in the table below. All objects are at rest. We have set the origin to be the position of object 1.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Can you please help me with finding the q without using sine, cosine, or tangent? Thank you!

Electric Force HW: Follow the Solutions Example and FAQ. Don't use angles. Two charged objects of the same mass (m = 1E[-3]kg) hang from a ceiling. The charge on each object is positive. Each object has the same amount of charge (q1 = q2 = +q). Each object is attached to a massless string of length L = 0.25m. The tops of each string are all attached at the same point on the ceiling. The position of each object is given in the table below. All objects are at rest. We have set the origin to be the position of object 1. The direction of each force can be found using the positions for the objects. For instance, the electric force on object 1 due to object 2 points in the same direction as a distance vector which points from the position of object 2 to object 1. You can find this distance vector by finding the difference between the position of object 1 (the final) and the position of object 2 (the initial). This distance vector is composed of a magnitude (distance) and a direction (unit vector). We only want the direction so that we can apply it to the force. We can find the 1 2 unit vector by dividing the distance vector by its magnitude (Pythagorean). a) Draw FBDS for each object b) Find all of the force directions (unit vectors). There are 3 for each object. c) Find the charge q and the magnitude of the tension in each string (T, and T2 ) Object х (m) у (m) 1 0.250 String Attach 0.125 0.217
Transcribed Image Text:Electric Force HW: Follow the Solutions Example and FAQ. Don't use angles. Two charged objects of the same mass (m = 1E[-3]kg) hang from a ceiling. The charge on each object is positive. Each object has the same amount of charge (q1 = q2 = +q). Each object is attached to a massless string of length L = 0.25m. The tops of each string are all attached at the same point on the ceiling. The position of each object is given in the table below. All objects are at rest. We have set the origin to be the position of object 1. The direction of each force can be found using the positions for the objects. For instance, the electric force on object 1 due to object 2 points in the same direction as a distance vector which points from the position of object 2 to object 1. You can find this distance vector by finding the difference between the position of object 1 (the final) and the position of object 2 (the initial). This distance vector is composed of a magnitude (distance) and a direction (unit vector). We only want the direction so that we can apply it to the force. We can find the 1 2 unit vector by dividing the distance vector by its magnitude (Pythagorean). a) Draw FBDS for each object b) Find all of the force directions (unit vectors). There are 3 for each object. c) Find the charge q and the magnitude of the tension in each string (T, and T2 ) Object х (m) у (m) 1 0.250 String Attach 0.125 0.217
e = 1.6E[-19] C Nm? k = 9E[9] C2 kąsqo -†so .2 Fet = F, + F2 + …* .. FE = q,E kąs Esp %| .2 ds
Transcribed Image Text:e = 1.6E[-19] C Nm? k = 9E[9] C2 kąsqo -†so .2 Fet = F, + F2 + …* .. FE = q,E kąs Esp %| .2 ds
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