(20 10? As shown in the figure, the distance between the vertically placed metal sheets M and N is D, and the left end of the insulated horizontal straight rod passes through the small hole in the cen

(20 10? As shown in the figure, the distance between the vertically placed metal sheets M and N is D, and the left end of the insulated horizontal straight rod passes through the small hole in the center of the N plate and contacts M.. (1) Because the P ball is positively charged, only when the potential of the M plate is higher than that of the N plate will the P ball be subjected to the electric field force to the right, and it is possible to move to the right.

Therefore, the ball P should be ejected from the plate along the horizontal straight bar, and the potential of the M plate must be higher than that of the N plate.

In order to make the ball p shoot out from the plate along the horizontal straight bar, the electric field force must be greater than the friction force on the ball, that is,

qE>μmg？ E=Ud。

So u > μ mgdq

(2) Let the velocity of P-ball when it emits electric field be V, which is obtained from the kinetic energy theorem:

Qu? μmgd= 12mv2？ 0

Namely. q5μmgdq？ μmgd= 12mv2

Solution: v = 22 μ GD

After the ball P is injected with a magnetic field, it is subjected to a vertical Lorentz force, and there are three possible situations.

I qvb = mg, then the ball moves in a straight line without friction, so the work done by friction is zero.

Ii. if qvb > mg, the straight rod will exert downward pressure on the ball, and the ball will slow down due to friction, and then move at a uniform speed when the Lorentz force is equal to gravity. Let the velocity vt of the ball be constant and qvtB=mg.

Let the work done by friction in this process be W, and we can get it from the kinetic energy theorem: W = 12 MVT2? 12mv2？

W = 12m (mgqb) 2？ 12m(22μgd)2 solution: W = M3G22Q2B2? 4mμgd

Ⅲ. If qvb < mg, the straight bar supports the ball upward, and the ball slows down due to friction, and the final speed drops to zero. The work done by friction is

W=0？ 12mv2

Solution: W=-4μmgd

Answer: The potential of M board is higher than that of N board, and the potential between them must be greater than μ mgdq; (2) The injection rate is 22μgd, and when qvB=mg, the work done by friction is zero.

When qvb > mg, the work done by friction is m3g22q2B2? 4 μ GD, when qvb < mg, the work done by friction is -4 μ GD.