Current location - Plastic Surgery and Aesthetics Network - Wedding supplies - (20 12? As shown in the figure, there is a uniform electric field with field strength E in the area above the horizontal line MN, and its direction refers to the upper right with MN sandwiched.
(20 12? As shown in the figure, there is a uniform electric field with field strength E in the area above the horizontal line MN, and its direction refers to the upper right with MN sandwiched.
(1) When negatively charged particles enter, they first make uniform deceleration motion, then make uniform acceleration motion in the opposite direction, and still enter the magnetic field for the second time from point O through MN at the speed of v0, as shown in the figure, which is obtained by Newton's second law.

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;