Question

Q2/ Let a Point charge Q1 = 25nC be located at p.(4, -2, 7)and a charge Q2 = 60nC be at p2(-3, 4,-2). In the free space of Cartesian coordinate system. Find: a) E at p3(1,2,3) b) E at the origin point. c) At what point on the y axis (0, y, 0) is E = 0.

Answer 1

Qut .P Solution given = 25 the nc located_at_2464,-2,7) and change Q2= 60mc_located_at_lz (-3,4-11 Q Note: Electric field vector due to a charge (Fp-ro). then electric field vector at at point P due to point charge des Ea = KQ (Fe-ra) 1 Ire-ro 13 here = 9x109 N-M2 4760 k= and to and re are the position vectors of charge aandpoint P with respect to origin respectively @) E at point B (1,2,3) w 10? P21 GQ2 Ta The total electric field vector at point_B3 will be Vai re 2) Corigin). here re-ral Ē = a + Ear +29+34)-(411-23 +7F) (92 (l rporal 37 + 4 2 - 4k

Date : - 41 and Tip-rail= 9 + 16 +16 Bernilarily (T-762)= (î +2ſ+ 3Â )-(-3 î+49-2 ) 4ix-2; +5 - and ir-razl= 16 + 4.725 45. thus 3 value. from 02 EQ, = KQ, (-31 +49-4R.) (640 Cetty similar substitute Eg = 9x109 x 25x179 (-31 +49 – 48) (14173 Eala 135 225 'I-3 i +49-4k] 3 (4123/2 aud similarily EQ2 こ + 5) KQ2 = (ai-2 (445)3 a z 9x109 x 60 X10 (41-29 +5€) (145)3 540 (41-29 + 5Ê (195) 4 Edi 3

Date : thus from ean (2). total electric field vector at point P3 es Ē = Eo = ERI + EQ2 Pog (3î +49-4K) + 540 (4-2) +5€) (4 crus13 c 225 3 0.86(-3i+45-48) + 1.79 (41 - 25 +516) (-2:58 1 + 8.443– 3.44 € ) + (7.16 – 3.58j+8.9518) 4.58 7 To (4563 0.143_+ 5051 ) NIC and magnitude is Erl= 7:16 NIC Answer. © Ē at origin ie as. 3 + EQ, (5) fp = 0,0,0) si Again__Electric field at origin can be written Eo Eat fromage ther rp-ro,= -41+2; -7 Eda Korea a lrp-ral 69

and Tp-rQ,= si-43 +2 29 ITp-TAI then from @ E = KQI (-41 +23°-7ť ) 9x10425x109 (-41+23-7€ (G93 (69)3 EQ = 0.39(-47' +29–7ť ) E - (-1.56 i + 0.78 j - 2.73€ and Ee2= KQ2 ( 31-4 + 2 € 9x109x60x169 (3 î-41 +26) (29) 3 (129)3 E Q2 = 8.46 ( 31-40+2€). a EQ2 = (10:38 13.84 ft 6.92 ) Trus from ean 6 FO: = EQ, +£ Eo = 8.8 (8.82 Ê - 13.06 + 4.19 ING and magnitude Eol= 16:31 NIC answer

-) zero © At what bointm of Haxis E=0 # here PuĚ (og,o) Total elefic electric field on y axis will be only if when the electric field due to individual charges. is equal anda_oposite direction le - En I Edi 1= Earl It mean their mangnitudies are same Note: then magnitude 8 Qo of electric field at charge point due to charge EQI point EQ= KQ ra 6 r= distance blue between charger and point- E Nom Now calculating settance between charge Qi and point Py. M = 16+04+22+49. riz 165+4j+272 and distance between point P és -charge Q2 and

Date : 12= g8+14-4)2 +4 82= 13 + 14-422 this. leoil Eool using egn 6 KQ, K.Q2 822 ㅋ QI ri? 2 822 72 of 25 x 10 9 65+ (4+2) Gox109 13+14=4² V 5 12 13+14-40? 65+18+272 =) 65+ 5142 +16–84) = 780 +.12142+4+44) =O 7y2 + 884 +683 the roots of this equation are 4,= -6.286 + 7.62 i 2 There are and P Y 2 = . -6.286 7.62 complex root answer