FORMAT A Required: (1 point) D. Substitution . ( 3 points) B Given: ( 1 point ) E Initial Answer: (1 point) C Formula ( 1 point) F Final Ansner (1 pont) DEDUCTIONS missing units (- 05 each ) not rounded to least SF (-0.5 each )PROBLEM SET 3. How far apart must two equal point charges of 75.0 x 10-9 C (typical of static electricity) be to have a force of 1, 80 N between them? 4. If two equal charges of 35.00 C each are separated in the air by a certain distance of 1.08x103 m, what is the magnitude of the force acting between them?PROBLEM SET 1. A force of 5.0 N is acting on the charge 6.08 x 18 C at any point. Determine the electric field intensity at that point. 2. What is the magnitude of a point charge whose electric field at a distance of 25.0 x 18 2 n is 3.40 N/C?PROBLEM SET 1. Find the flux surrounding a charge of 8. 85x10-12 C. 2. A circular plane, with a radius of 2.2 m, is immersed in an E-Field with a magnitude of 800 N/C. The field makes an angle of 20 with the plane. What is the magnitude of the flux through the plane? (Area of a cirle: A = 172)17 Required : Flux ($ ) Given , total charge , 9, = 8.85 * 1012 c Formula used . = Statal [Gauss's Theorem ) Co substitution Go stelative permitivity of face space Now 8.85 X15 2 Go 8. 85 X 1012 [ Go = 8. 85 *10 (2 N w? ] 4 = 1 Nm /c initial answer Hence the acquired flux is INm2/C . final answer Required : magnitude of the flux Grivers , readius of circular plate , R = 2.2m Electric field, E = 800 N /C angle , 0 = 20' 92 70 ( with plane ) . . angle with normal to the surface of= 90- 20" - 70 formula used = EAcosa ES electric field As area substitution Now Area of the plate , A = FIR 2 = 3.14 x 2.2 " = 3.14 X 2.2 x 2. 2 of A = 15 . 2 m 2 electric flux is given by = EAcost 9 = 1 800 x 15.2 x Cos 70' OF D = 80 0 X 15. 2 X 0 . 342 OY, 1 = 4158.72 Nm /c initial answell ) The value of sequined pluse is 4158.72 Nmi / c . ( final answer )5 % Required : distance between the charges , M=9 Given , : magnitude of equal charges , 9 , = 9 = 75 x 10 - Force between charges , f = IN formula used : F = _ 9192 4IJL GO J 2 [ Coulomb's law ] substitution NOW F Z LINGO 9192 41 2 ON 9 2 = 7 9X10 x 75X10 * $5x10 unGo = 9X10 Nm2/C 2 F 1 oy 9 X 75 x 75 X 10 - 82 = 50 . 625 X 10 - 6 V 30. 625X 106 = 7.12 x 10 on ( initial anneoen ) or, 4 = 7.12 X10 m Hence the required distance is 7.12 *10 -m (final answer) 4. > Required : magnitude of fone , F = ? Ciran , magnitude of equal charges, 9, = 9, = 35 C distance 91 = 1 x10 m formula used, F = 1 9192, UnGO substitution Now F = 9x10 X 35*35 9 x 10 * 35 X 35 = = 1. 1x 10 N initial ( 1 X 103 ) 2 1X106 F = 1.1X10 N Hence , the required force is 1.1x10 N ( final answer )1y Required : - Electric field intensity, E= ? Given Force, F = 5N Change 9 = 6X10 6 C formula used E - F/ 8 substitution Now E =F 2 5 6 X106 E = 8 33 X 10 N/ C initial andwort Hence, the required electric field ins 8. 33 x 105 N/C ( final answer21 Required, magnitude of change, 9 = ? Given Electric field E = 3.4 w/C distance , Me = 25 * 102. m Formula used E = 1 8 subshrubin putting the value 3.4 = 9 * 10 x 4 1= 9 *10 NW / C2 ( 25 X 10 2 ) 2 on 3 .4 = 9 x 10 x 9 625. X 104 2 ) 4 = 3-4 x 625 X10 9X 109 9 = 2.36 X 10" -1 C Initial answer Hence, the magnitude of change is 0-11 22 . 36 x 10 " C final