answer this with detailed solution sample solution below must follow neat and clean pleas

Equilateral Triangle B = 3 x 10 C 1409 A C = 4 x 10 CI 9 test charge = 6X 10 C : 5 X 10 C solve : - A to D - B to D - C to DXA SO - Resultant vector 9 PYONG nIST -qe = - 1.4 x 10- C 9 p = 1. 6 x 10 19 0 L 1350 x 10 N charge of positive charge 21011061 x 1072 the electron of the electron lod Nip13 0 0 1:1 4 POSITRON - 9+2 = 1350 x 10 = 1350, x 10 - 7+3N Problem 1 : = 1.350 x 10 # N or 0. 00 0 135 Two charges of Q 1; 10 x 10 c , Q 2 : 15 x 10 0 ebiollafam respectively are 10 cm apart . what is the force between two charges ? Problem 2: with vector quantities pronto out Given : Formula: d Nobs What is the total force exerted Q 1 = 10 x 10 - 9 c on the test charge If the three charges KQ1Q 2 71( forms a right triangle . With a Q 2 = 15 x 10- 9 C charge of 50 x 10 9 c at A , and r = 10 cm -> conversion of cm tom : 2405 x 10 - "C at C , and a test 4 Shortcut method : F = 1 0 0Ho 1 cm = 1 x 103 charge of 10 x 10- 9c. Find out what * = 10 cm m = 0 -m is the force of attraction of all - 2 + 1 - 2 add 0 r= 10 x 10 jon the charges of A to c to B r= 1.0 x 10 m 4 long method : 216 or 0. 1 F CB r = 10 cm x I'm 1 R / FAB 100 CAR 10 (ribs) slugmap of 150.137 sol'n: 0 240 B 1 Q B - 10 * 10 C LF = KQIQ2 r 2 53.130 to test charge 4 9 x 10 9 N - m ? 5 cm 10 * 10 - @ /15 x 10 @ : Q : 50 * 10 C 3cm (1.0 x 10 ' m ) 2 L (9) ( 10 ) (15 ) x 10 N - m/ fir A C QC = 25 x 10 9 c -4 cm 1.0 x 10-2 1/2 Sin A =. 3 5 297 + 2B = 90 - A sin A = 0- 6 - 90 - 34.870 A = 36. 8701. FAB DISTANCE FROM A TOB 3. RESULTANT VECTOR - taken by COSINLAW FAB KQAQB R = J ( FAB ) 2 + ( F BC ) 2 + 2 FAB FBC COS 53. 131 ( FAB ) 2 R = 0 . 00 3 146 or rAB = 5 cm m = 0.05 m or 5 x 10 m = 3. 86 x 10 - 3 4 = ( 9 x 10 9 N - m 2 ) (50 * 10 # @ ) ( 10 x 10 ; ) cancel units that (5 x 10 - 2 m ) 2 are the same 4 = (9) ( 5 6 ) ( 10 ) x 10 9 - 9 - 9 N-m/2 Formula for this : 35 x 10-4 m z R = V ( Z FAB ) 2 + (