The distance between two horizontally placed metal plates is d

Between the two plates, there is a positively charged oil droplet with mass m and charge Q that happens to be at rest, indicating that the upper plate is negatively charged and the lower plate is positively charged.

The top of the resistor is equivalent to the negative pole, the bottom is the positive pole, and the current is from the bottom to the top, that is, the magnetic field generated by the induced current is n on the upper side, and the magnetic induction intensity of the induced current is straight down.

a Magnetic induction intensity b is vertically upward and is increasing The magnetic field of the induced current hinders the enhancement of the magnetic field The magnetic induction intensity of the induced current generated is in the opposite direction to the direction of the original magnetic field and is vertically downward.

d The magnetic induction intensity b is vertically downward and is weakening, and the magnetic field of the induced current hinders the weakening of the magnetic field, and the magnetic induction intensity of the induced current generated is in the same direction as that of the original magnetic field, vertically downward.

So you can only choose between A and D.

The answer you wrote is not correct, and the above should be divided.

qr）d=mg

δt＝mgd(r＋r)/nqr

So D should be chosen

The static oil droplet indicates that the lower plate of the capacitor is positively charged, and the current in the coil is from top to bottom (inside the power supply), and it can be judged by Lenz's law that the magnetic induction intensity b in the coil is an upward weakening or downward enhancement

and e=nδ δ t

U*r(r is the subscript)=[r (r+r)]*e q*ur(r is the subscript)] d=mg

Obtained by:δ δ t=[mgd(r+r)} nrqAnswer: c

In 0s 10 -4s, the particles are subjected to the opposite direction of the Lorentz force and the electric field force at the same time, their magnitude is fb=qv0b=5*10 -7, fe=qu d=5*10 -7, they are equal in size and opposite in the direction of the resultant force is zero, do uniform linear motion, in 10 -4 2*10 -4s, the particles are only subjected to the Lorentz force, make a uniform circle, the orbital radius is r=mv0 qb=, the period t=1*10 -4s, so the particles first move horizontally to the right for half a cycle, Then do a uniform circular motion, and just do a week, repeat this motion, because of the radius r

Under the action of the Lorenz force, the electrons deflect downward and move in a circular motion. The maximum deflection is the uppermost electron, and the uppermost electron is deflected by 5d, which just can't be shot out between the two plates

Let the radius r of the circle, which is solved by the geometric relation: r -(r-d) =25d r=13d by r=mv0 eb=13d: b=mv0 13ed, which is the minimum b

1）qu=mv0^2/2

U=mv0 2 2Q= *(4 10 7) 2 2*2) is the maximum displacement at 10 cm from the center of the phosphor screen.

The time of movement of electrons in the electric field t=l v0=

Let the voltage applied between the two plates be U', the acceleration of electrons in the electric field is.

a=qu'/md

The deflection displacement of electrons in the electric field y=at2 2=qu'md*t 2 2 From the flat containing this throwing motion law, it can be seen that the electrons move in a straight line along the velocity direction after leaving the electric field, and hit the phosphor screen at the point p. From the inscription, when the P point is 10cm away from the center of the phosphor screen, the plate voltage is the maximum, and it is set to U'。

It is also known from the law of flat throwing motion that the straight line along the direction of velocity intersects the center line of the parallel plate at the midpoint.

From the similarity of triangles, l 2:(l 2+l2)=

y=Substituting the previous relation.

md*t^2/2

u'= In order for the charged particles to hit all positions of the phosphor screen, the range of voltage applied between the two plates should be u'<364v

The answer is suspected of judging the lack of teasing see the figure below, if you are not clear, you can ask Qin to change!

Find the source Zheng Kutu in the hail hole in Douding, which is only for reference to the cong to investigate.

The conclusion is that if the upper plate is moved, the motion of the particle does not change, and the particle can still reach the n hole and return the same way. If the lower plate is moved, the upper movement will make the particle return at a certain point between the plates on the N-hole, and the downward movement will make the particle continue to move downward through the n-hole.

The answer is ACD

This is a question about the functional relationship, and the particle point can just reach the n hole, which shows that the positive work done by gravity is equal to the work of overcoming the electric field force, and the work of the resultant force is equal to zero.

If plate A is moved, the work done by gravity does not change when the particle reaches n, the voltage between AB does not change, and the work of the electric field force does not change, so the kinetic energy of the particle is also zero, and then returns, so A is right and B is wrong.

If the B plate is moved upward, the work done by gravity decreases when the particle reaches n, the voltage between ab does not change, and the work of the electric field force does not change, so the kinetic energy of the particle becomes zero before reaching n, and then returns, so c is right.

If the B plate is moved downward, the work done by gravity increases when the particle reaches N, the voltage between AB does not change, and the work of the electric field force does not change, so the kinetic energy of the particle cannot be zero when it reaches N, and it will only continue to fall by gravity after passing N, so D is right.

Solution: According to the problem: mg(d1+d2)=eqd 2 is obtained from e=u d2 mg(d1+d2)=uq

Regardless of whether plate A moves up or down, the work done by the electric field is certain, and the work done by gravity is also certain, so A is correct and B is wrong;

Suppose that the particle can reach the n-port;

b plate moves up, d2 decreases, mg(d1+d2)uq, there must be a moment in front, mg(d1+d2i)=uq, equal to the back, because gravity is greater than the electric field force, the particle will continue to fall, so d is correct;

Select ACD

When it comes to the particle point, gravity is not considered, the start of falling is acceleration, after passing through the A plate, the two metal plates all produce resistance, start to decelerate, and stop when to N, if A moves up, then the acceleration becomes shorter, but the speed to M becomes smaller, and after deceleration after M, it can still return, A is correct, A moves downward, the acceleration time becomes longer, and the deceleration becomes shorter after M, so there is still speed after N, and it will continue to fall, B is correct, and CD is wrong

Rough explanation、I hope I understand、