x C User Dashboa x M Your question x Microcoulomb x MG sin cos tan G x User Dash

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x C User Dashboa x M Your question x Microcoulomb x MG sin cos tan G x User Dashboa x Dashboard To x User Dashboa x VA Hw1 Sabella C secure https://www.webassign.net web/Student/Assignment-Responses/submit?dep 15369258 3. 0/2 points I Previous Answers CJ9 18.P 013. My Notes Ask Your Teacher Two point charges are fixed on the y axis: a negative point charge q 28 HC at y1 +0.23 m and a positive point charge q2 at y2 +0.31 m. A third point charge q +8.1 HC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 21 N and points in the ty direction. Determine the magnitude of q2. 7.85 10**-5 X C 4. 1/2 points I Previous Answers CJ9 18.P021 My Notes Ask Your Teacher Interactive LearningWare 18.2 provides one approach to solving problems such as this one. The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1 +8.55 HC; the other two charges have identical magnitudes, but opposite signs: q2 4.50 HC and q3 +4.50 HC. 230e 23.0? ,30 n (a) Determine the net force (magnitude and direction) exerted on q1 by the other two charges. magnitude 25 direction along the +y axis (b) If q1 had a mass of 1.65 g and it were free to move, what would be its acceleration? magnitude 5151 X m/s rection along the +y axis Additional Materials H Video javascript:openWindow('/CJ/interactive learningwarelhtm/lnteractive Learningware 18.2.htm, teractive learningware', "width 700,height 500,scrollbars-yes,resizable-yes,menubar yes

Explanation / Answer

3) 9*10^9*28*10^-6*8.1*10^-6/0.23^2 = 38.59 N +Y

9*10^9*q2*8.1*10^-6/0.31^2= 758584.81 q2 -Y

38.59 - 758584.81 q2= 21

q2= 23.19 *10^-6 C = 2.32*10^-5 C

4) F= Kqq/r^2 = 9*10^9*8.55*10^-6*4.50*10^-6/1.3^2= 0.2049 N

Net force = 2* 0.2049 sin23 = 0.16 N

a= F/m = 0.16*1000/1.65= 97.04 m/s^2

check the work once

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