Hi I cannot figure out part a and b. Thanks. MasterinqPhysics Homework 2 on elec
ID: 1657291 • Letter: H
Question
Hi I cannot figure out part a and b. Thanks.
MasterinqPhysics Homework 2 on electric fields and electnc potential Mozlla Firetox https://session.masteringphysics.com/myct/itemView?assignmentProblemID=82905124 0 ields adlProblem 23.47 Resource3 v a previnus 29 ot 3 next Problem 23.47 Part A A point charge 390 nC point charge 12--3 10 nC s placed at the ongin, and a second placed on the z-8xis alz-+ 2 1 0 cm What is the potential energy of the system of the three charges if gis placed at 10.5 cm? A third point charge -1.95 nC is lo be placed on the z-axis between 91 and2 (Take as zero the potential energy of the three charges when they are ininiey far aparl) U1.22700114.10 Submit M Answer Give Up Incorrect; Try Again; 3 attempts remaining Part B Where should g be placed between gi and q2 to make the potential energy of the system equal to zero? Submit My Answers Give Up de Feedba 1245 PM 9/23/2017 Type here to search "Explanation / Answer
a)
Here, we have three smaller systems,q1 and q2,q2 and q3 and q1and q3. Each system has its own potential energy, and the three potential energies add up to form the potential energy of the bigger system.
UT = U12 + U23 + U13
UT = kq1q2 / (r12) + kq2q3 / (r23) + kq1q3 / (r13)
UT = (9 x 10^9)[(3.90 x 10^-9 x -3.10 x 10^-9) / (0.21) + (-3.10 x 10^-9 x 1.95 x 10^-9) / (0.105) + (3.90 x 10^-9 x 1.95 x 10^-9) / (0.105) ]
UT = -3.84 x 10^-6 J
b)
Let the charge q3 be placed at an arbitrary point x along the x -axis. Note that the problem states that q3 lies between q1 and q2, thus 0 < x < 21 cm
UT = U12 + U23 + U13
UT = kq1q2 / (r12)^2 + kq2q3 / (r23)^2 + kq1q3 / (r13)^2
UT = (9 x 10^9)[(3.90 x 10^-9 x -3.10 x 10^-9) / (0.21 x 10^-2) + (-3.10 x 10^-9 x 1.95 x 10^-9) / (21 - x) x 10^-2 + (3.90 x 10^-9 x 1.95 x 10^-9) / (x x 10^-2)^2 ]
0 = -12.09 / 21 - 6.045 / (21 - x) + 7.605 / x
- 6.045 / (21 - x) + 7.605 / x = 0.5757
(159.705 - 13.65x) / (21x - x^2) = 0.5757
(159.705 - 13.65x) = 12.0897 - 0.5757x^2
0.5757x^2 - 13.65x + 147.6 = 0
Please solve the quadratic equation...