Question

1 pts Question 3 0.5 mas shown in the figure. The magnitude of the electric field at the Four point charges (q - 15C) are fixed in position at the corners of a square of side a center of the square is 12x10 NC 31x10NC 3.4x10NC 9x10NC 73NC

Answer 1

029) A 45 M€. = 39 4 point changes are situated at A, B, C, D respectively a at By 60 M = 49 at ca 15 M & q. n d A (39) B 49) at D 30 MC = 29. - We've to evahate Resullaut field at the center of the - square of side AB=0.5m. 1 m Now, OA = OB = OG 2 OD . ? & Vet vej utram. - Now, field at o due chaye Bis – L. 60.0106 - 9x 180 x 60 x 18 x 8 vim Um = 432 xyo voi here, 1N/C ? V/m 24-32 X 106 N/C along of - -- h. 30 room m. 9x10 30x10x8 Sinilow field at o. due to chaye Dis - Unto. 7/212) 2 -2.16 *106 N/Ç. along. Therefore, the resultant electric field due to branje B&D is = (4.32.2-16) x103 view – 2.16 x 168 von along oD. Similarly, the wesultant electric Steld dus to chage Aue 4 us10 - is unto, 1/2+2) 2 15 x 18 UNG, 1/2 22 along along oc - 9x109 x 8 x (45-15) x 18 = 2:16 x 108 Vow alery one! Therefore, the resultant electric field due to the draxer at AB, C, D is along ok a V (2.16) ² + (2.16) 2 x 180 von here INIC = I vlony = 3-054 X10 Vlow alory of

of Here oo & OD that is the resultant direction fields due to change (Ac) & (BD) nespectively electric are at net resultant electric field is along o direction. 2. Of = Vocyt@D?+2.0C.0D.COS 90° a pochi codje - = 3-054x10 Nic Here in this giren McQ Question , in the five options given Option @ is correct, ie. I = 3.1X100 N/C - B = 31 XIÓ NIC