Question

Problem-2 A very thin, uniformly-charged rod with total charge Q and length 2L lies along the vertical 2-axis. Sitting still (zero velocity and zero acceleration) at a distance H above the top end of the rod is a proton. Gravity is acting on the proton in the downward (-2) direction. What is the the total charge Q on the rod? + profon K & de d = k l . dx. - EL - K Let =) , U, - X L us Yo-x du = – dx Xoth - du ut XO-L Lo + 2 F = 10L 2 L - Q = E 2L ( - 6 subtract , murupy - To convert Celsius (Cº) to Fahrenheit (F°): multiply by 9, divide by 5 and then add 32. 810T TUL 9 9 18 27 36 45 54 63 72 81 30 10 10 20 30 40 50 60 70 80 90 100 11 11 22 33 44 55 66 77 88 99 110 12 12 24 36 48 60 72 84 96 108 120 110 1 121 1 132

Answer 1

P & . Electric field at a distance it from a uniformly charged rod of length 2L & total charge &. lineer charge density of nood, a-1 charge on the line element dx ip dq= idx field at P due to that line element is, de - o dat die So, total field at Pir, x=4+24 E =Sde = 1 Arte [- < AC L +21 to [ +421 - 4 A B x=H60 5+2 L H+2L - HITE 4TGO

- A - E= 401 G H-H-2L (H+24) 211 +(H+24) - 1 4 ITG 9 => E- a + (8 +21 I electric field at P if the man of the proton ip mp & charge ip e then upward electric force acting on it ip F = e E = 1 e 9 I ALTE H(H+2L) & downward gravitational force acting on it ip, Wome go F Sr pautans ip zow] WoF , as the net force on mp g = 450 H (H +2 L) 41TE0 mp G H (H + 2L)