Question

Three charges are located along y-axis with first -50 C charge at origin, second -100 C charge 2m above origin and third +50 C charge 3m below origin. (a) What is the magnitude and direction of force exerted by second charge on first charge, (b) what is the magnitude and direction of force exerted by third charge on first charge, (c) What is the net force on first charge? (d) If the third charge is replaced by an imaginary test charge find the magnitude and direction of electric field at its location due to first and second charges.

please show steps of breakdown to each problem

Answer 1

3 m 2m 13 d 12. T23 92 93 93E = t 50c 2 r Im Given duta :- 유 = -50C 2 100C - 2m, K3 = 3m 23 Force exerted by second charge on first charge we will denote it by F 12 ܐܩܓ 2.1 F 92 X9u r3 4TTE. K 9291 r3 12 2 9 x 109 x (-10oC) x ( -50 c) (2) 9 x 109 x 5000 4 = || 250 X109 13 1. 125 x 10 Fax N ( magnitude) Since q is-ve, q is-ve, hence fórce is repulsive. Hence, direction is away from second charge Hence Direction is ve x-axis food Force exerted ie by ch third 93 891 charge st charge 1 F 31 4TTED (13 2

Eziz x x (+50) x(-50) (3) 9 x 109 x (- 2500) g 31 - 2500 x 109 2.5 X1012 N (magnitude) Since gis tve, q, is-ve, hence force is attractive Direction is tve x - axis le force on Net first charge, we have 1.125 x 1013_N 11.25 X1012 N & direction is -ve x =-11.25 x 1012 î - 21 Ē N 21 Similarly = 2.5 X 1012 î N E 31 ETRE E F 2! E E iu. U Net 31 il 771 2.5 X10 2 11.25 X1012 î 8.75 X 10'2 în Net INCE Magnitude: 3.75 X1012 N direction =_=X_direction (d) __Electric field at point of a

Electric field due to 9 at point of go E = k 91 Ta 9x 109 x (-50) (3) -50 X 109 NC Ę = -5 X1010 î NLC Similarly, electric field due to ta at 3 E K 92 1233 9 x 109 x 1-100) (02 -900 X 109 N/C =-90x1010 1 N/C Total electric field due to ENet charge and 유 2 - 5X 100 gox Iolo -95 xlolo î NLC E Net -9.5 X10" î NE Magnitude 9.5 X10" NIC Direction -Ve X - axis.