Question

1. Three point charges are located along the x axis as follows:  at Q1=30nC at (0,0,0), Q2=-27nC at (12,0,0) and Q3 at (x,0,0) (all locations given on meters). Solve the following scenarios:
   1. Considering charge Q3=10nC , what is the total force applied to this charge if x=5 m.
   2. If the charge Q3=10nC , what must be the distance X so that the total force applied to the charge 3 is equal to Ft=-288.80nN .

1. Using the three same charges from previous problem located at Q1(0,0,0), Q2(3, 5, -2) and Q3(1,2,3).  What is the total electric field at points:
   1. P(1,1,1)
   2. M(-2,4,-6)

Answer 1

0 - ܝܢ AR 47 Eo QR- FLD (30x19 toxicy i sam Due to point charge the electric force, Ē IR12 → unit vector Corresponding to ē Qi = 30ne, c9,90) Q2 = -27nc, (12,0,0) Q3 = lonc, (5,0,0) (A) total force ON Q3. F= Fit F2 Q1.03. år IR12 [(5,0,0), ( 1,0,0) We calculated force on charge Q3 dueto Qq Q is final Point Ř = sån R1=5 (512 9X 18 X 3 X 1 XXIXIE fis F = 1 47{o R = so 45{o A a - 25 A F = 108.8109 Nam CS scanned with CamScanner

- F2= Q3.02 (OR) 2) 45EU R = 1 R12 [ (5,0,0) -( 12,0,0)] Ř = -7a 11 = 1 (quxlc 9 (-29x109) F2 = (72 7an 7 4TEO 9x18x qoxi* X (-27 X117) [- .au F2= 49 an F2 = 7249-33810 an Ft= FI+ Fz Ft= (108 X107+049.33 Xing) Fy= 156.77 8159 Nam Fy= 257.77 kN an A Fr= 157.33 han (B) Force on charge az 'due to other two charge is -288.8 nn , then find of a F= Fit F2 Rs.22. CARD 42L. 1812 Ř = [0,0,0) [0,0,0] FL- Ą Ř = xan Scanned with CamScanner

nañ 3 F= 10X16?x 30X16*Pax 4TE (x2 [them an F = 9x1pMx 10x10 x 30x109 x2 - ^ 2700X16 an ILI 22 F = Q3.18 Q3.12 ă 4ā Eol R12 R= [ (2,0,0), (12,0,0)] au R = (2-12) an 20x157x(-27x19 ai ] با F2 47E (2-122 an E = 3x1wxloxiXXX 1-23 X109) (1-12; 2 an F2 = - 2430 X10 (2-122 FT = FI Fit F2 2700 8109 -243 0 Xiing 92 + 22 (2-122 CS Scanned with CamScanner

FI= - 288.8x15 AM 4 Ro 2700 -2 ११.१४॥ २५30 (1-12)2 - -288.8 2700 2430 (A-12)2 M- 2100 + 288 - २५30 (x-12) २५30 भ+५५ - २५४ 200+ 2१४.१ - Pr २५303 (2300+28.2) (+५५-२५४) २५30- 2900 + 2.१४. १४१ + 3 १११०० + +1587.22 64800 6937 203 -412592= २४.21- (१31.283 - ५१००४-+ 3 2 220b २ ११.४५ - ६११.2x + १.१ 572-- 6५४ 0 0 + 3 ???007 We solve. And get a value. CS Scanned with CamScanner

5) Electric freed due to point charge Q AR Ee 47Eo 1 R12 (A) E. at put,i,1) due to a(0,0,0) (30nc) 02 (3,5,-2) Q3 (1, 2, 3) E = E,+E2+ E3 E,= Qi GR 4560 IR 12 ЭС-2}nc) y (lone) Ř = [(1,1,1),(0,0,0)] R antag+ 93 IR = 3 9x108x30x18 an +92 +9 Eia sve to say ) 9x108xx 343 antagta E1 E,= 270 [an 192 +93 33 57.9694+ 51.96 + 51.9699 1 A VIM - CS Scarried with CamScanner E

(6 9 Ez 9R Eas fQ 2 4Ats (| -23X1 x 9X107 & I- [1,1,1) [ 3,5,-2] 2 = -299 -49% +3 4 8| =439 ++ 16+9 R R - 23x9 E2 = [ 24 - 43 + 38 २५3 E2= C J31 (-394 - +34 ] -3 | 4 – 6:23 9% + 4 แห่ง V/n (เ: 6 ค E 94- Ez a 3 12 ( 8 ) | 98 | | E 1sxu x 3x4 (R / 9 | 8 | E = A R = (1, 1, 1) (1,2,3) K -44 - 25 (8)=N CS Scanned with CamScanner

ED = 90 E= 51.966 + 51.9699 +5190 -- 3-11 99-6-2243 (V5) 2 Es E= Et fit √5 [ aģ–2a3 ] 8.04 – 16.0893 V/m +4-6693 -- 8.04 ag - 16.084 +37.792 +40.54 @z Vim ANS ES also 48.85mm similary we can calculated at point [°~2, 4,-6] also sxvo} CS Scanned with CamScanner