QE=ma and a=qEm.
Let the time of uniform deceleration motion of particles be t 1, and the reverse uniform acceleration motion be t2. According to the symmetry of motion,
T 1=t2=v0a Yes.
t=t 1+t2=2v0a=2mv0qE
(2) Let the radius of uniform circular motion of particles in the magnetic field be r, which is obtained by Newton's second law.
Qv0B=mv20R，R=mv0qB。
The position of the particle passing through MN for the third time is c, which is obtained from the geometric relationship?
s=2Rsinθ=2mv0qB
(3) When the particles enter the electric field from point C, they make quasi-flat throwing motion, and the position where they pass MN for the fourth time is d, the included angle between the CD direction of the particles at point C and the velocity v0 direction is 45, and the time from point C to point D is t3, then
If a particle perpendicular to the e direction moves in a straight line at a uniform speed, what will happen? s 1=v0t3
Do a uniformly accelerated linear motion parallel to the E direction, and the initial velocity is zero, will there be? s2= 12at23=qE2mt23
Tan45 =s2s 1+0。
Simultaneous solution? t3=2mv0qE
The velocity perpendicular to the e direction is v 1=v0.
What is the speed parallel to the e direction? v2=at3=qEm？ 2mv0qE=2v0
So when the particle passes MN for the fourth time, the velocity v is v= praise and trample. What is your evaluation of this answer? Put away comments//high quality or satisfaction or special types or recommended answers dottimewindow.iperformance & window.iperformance.mark ('c _ best',+newdate); Lawyer recommendation service: If your problem has not been solved, please describe your problem in detail and consult other similar problems free of charge through Baidu Law Pro 20 12-03-05. As shown in the figure, there is a vertical uniform electric field below the dotted line MN, with the field strength E=2000v/m, and there is a vertical position 1020 16-07 above the electric field area. As shown in the figure, there is a uniform electric field with field strength E on the left side of MN, and the electric field is downward and parallel to MN. There is a vertical piece of paper 220 14-02-07 on the right side of Mn. As shown in figure (a), there is a large enough uniform electric field in the xoy plane. The direction of the electric field is vertical and upward, and the electric field intensity E = 40N/C, and there is 22065438 on the left side plane of the Y axis. Mianyang simulation) As shown in the figure, a uniform electric field E and a uniform magnetic field B are perpendicular to each other in a certain area with the width l=0.8m, and the size E = 2×108n/c62012-06-19. As shown in the figure, the horizontal uniform electric field strength is E (the field width is L, and the vertical direction is long enough. Next to the electric field is 220 12-06- 17, which is vertical and straight outward. As shown in the figure, the field strength of the horizontal uniform electric field is E (the width of the field is L, and the vertical direction is long enough), and next to the electric field is112016-09-04, which is vertical and outward. In area I, there is a uniform electric field E 1 with a 45-degree angle with a width of d 1. There are orthogonal bounded uniform magnetic fields b and 320 16-08- 19 in zone II, and there is a uniform electric field with a 30-degree angle above the horizontal line MN. I recommend the electric field strength to you, especially F.context('cmsRight', [{'URL':'/d 01373f082025aaf51kloc-0/aa256e9edab64034f1a07? x-BCE-process = image % 2f resize % 2Cm _ lfit % 2Cw _ 450% 2Ch _ 600% 2c limit _ 1% 2f quality % 2Cq _ 85% 2f format % 2Cf _ auto '，' contractId':'A24KA00562 '，}])； The price of electric cars has dropped many times. Is the quality guaranteed? What impact will the "network toilet" have? Is Huaqiang North's second-hand mobile phone reliable? Why is the cost of cancer treatment getting higher and higher? I recommend f.context ('recbrand', [{"img": "\/86D6277F9E2F07083523F69DFB 24B 899A901F20d? x-BCE-process = image % 2f resize % 2Cm _ lfit % 2Cw _ 450% 2Ch _ 600% 2c limit _ 1% 2f quality % 2Cq _ 85% 2f format % 2Cf _ auto "，" url":"/hm.js？ 6859 ce 5a af 00 FB 00387 e 6434 E4 FCC 925 "； var s = document . getelementsbytagname(" script ")[0]； s.parentNode.insertBefore(hm，s)； })(); window . TT = 172 134 1853；