Question

In a right angle traingle ABC, angle ABC is 90 Degree, AB = 2 m, and angle ACB is 41.81 Degree. A point charge of 5*38 nC is placed at point C, point charge 4* 38 nC is placed at point A and point charge 1 C is placed in point B. Calculate the force on charge at B due to others two.

Answer 1

Answer 02 W LE Given, AB = 2m LABC = 90° LACB = 0 = 41.810 charge at point cis, q = 5.38 nc = 5.38×100 at point B is, q2 = AC charge charge at point is, q3 = 4.38 nc = 4.28X10°C Now, Be Coto - BC = A B Coto Bc = 2 x Cot 41.810 m BC = 2.236 m Now, force on 93 due to charge q, at c is Fit 993 ATTto (Ber along en

9x109 x 5.38*109 x 1 (2.2362 Balong CB and, Force on 92 9.68 N, along en due to charge az at A is F = 1 Q2B Aito ; along An (AB) 1x 4.38×10 = F = 9x 109 x Nj along AB F2 = 9.86 N ; along here, angle between F and F2 is gou Therefore magnitude of net fored F = JFT + + 2F1 F2 Cos 90° (9.68)+ 19.86)70 N as 60sg0°=( I =13.82N

Now, if o be the angle between F and 0,= tan! ( 2 ) Ĥ then 0.86 .68 = 45.53° Now, if & be the angle off with the Be x = 1800+ 45.53 x= 225.53° then, of force on B Therefore the magnitude due to the two others charge is F = 13.82N will be The direction of force along the time makes an angle 225.530 with the line Bc in the counter-clock wise direction .