Question

triangle. We shall consider the process of the work required to assemble this collection of charges in the manner shown. To do so, we shall start with empty space and then add into that space one charge at a time until we have completely assembled the triangle of three charges. The legs of the right triangle have dimensions dx = 6.16e-02 meters and dy = 1.1e-02 meters. de 92 93 dy 91 The charges are as follows: • 91 = -7.44-06 Coulombs. • 92 = 1.67e-06 Coulombs. 93 = -5.77e-06 Coulombs. Determine the work required to place only 91 into the position shown: Joules Determine the work required to ADD 42 into the picture: Joules Determine the work required to ADD 43 into the picture: Joules Determine the total work required to complete all the above steps: Joules NOTE: Some of the works are positive. Some of the works are negative. Keep LOTS of digits if you want to get the sum correct to the required precision. Submit Answer

Answer 1

only 2 is 2 = 106766 -6 C 23= dr = (a) Only '&' Wi=0 when Solution First when placed at it's position, there -5.77410 no other charge henre incoming charge 2 will not feel 6.16 210-02 any potential, hence No Idy=1011-02 work will be done So work done to place 2,2-7040 b c into the position shown in Fero Now 2 = 1067410-be is added at it's posilos it will be under the effect ob already present charge 2 q, 22 [ 188=14103 Nm2/C² dy W₂ = =-1010111x 10 jould 1.1x10-2 W22 - 1001111 joules when ca is added, it will be in the presence obr, and a neet the in W2 = unco 9x109 x 704 & polo x 10674166 Now work done i Wz- ㅗ UTTEO + de + dy? 6 10-2 こ dn =9x100x-5.77210 210-6 (1-67-704 6.16 6.25744 47.33354 10 1 Joules = 4.733359 joules 4.7334 Joules Total work done W, +W₂ TW3 =5.377747 Joules - – 5.378 joules & d