3. There is a distribution of charges as evidenced below where q1 = +5.0 nC and q2 = -3.0 nC. If a third charge of magnitude 2.00 nC is placed between the first two charges at a distance of 15.00 cm from the origin along the x-axis, what is the potential energy of the system? You may assume that the third charge originated from an infinite distance away. Now if we wanted the potential energy of the system to be zero, where should the third charge be placed (hint: you will need to simplify and then solve a quadratic or you can use solver in excel)? Q3 X x=15.00 cm x=30.00 cmQuestion 4 6.5 / 6.5 pts From Problem 3 There is a distribution of charges . what is the potential energy of the system? Give the answer in units of uJ. -0.2 Question 5 6 / 6 pts From Problem 3 There is a distribution of charges .. Now if we wanted the potential energy of the system to be zero, where should the third charge be placed ? Give the answer in units of cm from the origin along the x-axis. 12Sorry I misread the question. ( a) Given, 2 . = 5hC 92 = - 3 nc 23 = 2ne Total potential energy = 1 9193 + *9392 + 1 9, 92 8 2 = KX10-18 5X2 + 3x(-2) + 5(-3) 0.15 0.15 0 . 3 = 9X109 X 10-18 20 12 0 . 2 = 3 x10-8 [- 7] = - 21 X10-8 = - 0. 2 MJ (b) Let the change 93 be placed at distance a from 91 Total potential = K 9192 + K 9193 + K9392 energy 0.3 x 0 .3 - X = KX10-18 5 (- 3 ) + 5 X 2 + 31-2) 0 - 3 0 .3 - X = 0 - 15 + 10 6 0 . 3 X 0 . 3 - x = 0 - 50 ( 0.32- 712) + 10( 0.3-21 ) - 62 = 0 = 50x2- 15 x + 3- 102 - 621 =0 = 50 x- - 31x + 3 = 0 * = 31+ 312-4(50)(3) 2 (100) = 01 12 or 0.5 ( not possible ) :. It should be placed at 10.12 m from 9, in positive -apis