For electric field science problems, ask a physics master to get extra points

Analysis of movements.

In the horizontal direction, it is only affected by the electric field force.

u d=e then there is Newton's theorem of motion.

eq=ma1

d=1/2a1*t^2

where a1 is the acceleration in the horizontal direction.

You can find the expression for t.

t=√（2eq/m）

Vertically, the falling process can be seen as a free fall.

So. h=1/2g(t1)^2

Because the vertical direction of the ascent and the falling process is symmetrical, then.

t1=1/2t

Yes. h=1/4·eqg/dm

Then the speed of the up-toss.

v=gt1=√（2uq/dm）·g/2

It is necessary to analyze the forces from both directions.

and both horizontal and vertical directions.

The horizontal direction is only a horizontal motion with an initial velocity of 0 due to the gravitational force of the electric field f(eq).

The vertical direction can be seen as a free-fall motion.

Only gravity mg acts.

Let's divide it. Let's do it directly.

The motion of the ball can be regarded as a uniform acceleration linear motion in the horizontal direction and an upward throwing motion in the vertical direction.

Horizontal: The force is the electric field force qe=qu d, so the acceleration a=qu (dm), write the equation of motion.

d=(1/2)*a*t^2 (1)

Vertical direction: the force is gravity, so the acceleration is g, because the ball returns to the same height, the displacement is 0, and the equation of motion is written.

0=v0*t-(1 2)*g*t 2 (2) is solved by equation (2): t=2*v0 g, and substituted by equation (1) obtains:

v0=d*g*(m/2*q*u)^(1/2)

+ a___

b You may wish to set a parallel plate as shown in the above figure, because m flies horizontally at the edge of a and can fly to meet n, m is the positive charge q; n Fly in horizontally from the ** horizontal line, which may move upwards or downwards; To ensure the encounter, the horizontal velocity must be equal, and the horizontal direction does not need to be considered, which can be equivalent to the encounter of two objects in the vertical direction.

i) Consider the upward motion of the N parabola, i.e., N is negatively charged, in this case, Mn moves in opposite directions, and as long as the plates are long enough, Mn will meet.

ii) If n is also positively charged, and the magnitude p, then the acceleration of n must be less than the acceleration of m to satisfy the condition of the encounter (m chases n), and since m and n have equal mass, the charge of m must be greater than the charge of n.

Summary: The main premise - m and n must enter the electric field at equal horizontal velocities in order for m and n to meet.

1) If m and n have the same kind of charge, the charge of m must be greater than the charge of n, the acceleration of m is greater than the acceleration of n, the two charges enter the electric field until the two charges meet, and the work done by the electric field force on the charge m is greater than the work done by the electric field force on the charge n;

2) If m and n have a different charge, m and n will meet under the premise (in any case).

The correct answer is only d

a=eq m, vertical displacement x=(at 2) 2 from simultaneous motion to meeting, t is equal.

So from xm>xn, am>an

Since e is equal, the specific charge of m is greater than the specific charge of n.

The charge is like a flat throwing motion, there is Y 12uqdmt2 in the vertical direction, because the time, voltage, and plate spacing are equal, so the specific charge Qm of charge m is larger, and the amount of charge is also larger, A is correct.

There are no figures. Are they both going in one direction, or are they going in different directions?

However, if they were entering the electric field in the same direction at that time, d must be correct, otherwise, they would not have passed the same position at the same time in the horizontal direction and would not have met.

9 4 3 (or.)

When the switch is disconnected, the resistance of the external circuit is equivalent to two 18 ohm resistors in parallel, which is 9 ohms, according to Ohm's law of the closed circuit, the voltage at the end of the road is and the voltage at both ends of R2 is, according to the principle of voltage division, the voltage at both ends of R1 is 9V, and the voltage at both ends of R2 is 3V, and indirectly in the two resistors, the potential of God carries zero, then the potential of point A is +9V, and the potential of point B is.

Then the potential at the upper end of R3 is also 9V, and the potential at the lower end of R4 is also 9V. According to the principle of partial voltage defeat, R3 partial voltage, R4 partial voltage 9V, then the electric potential at point D is or.

When the switch is closed, the total resistor of R1 and R3 in parallel is connected in series with R2 and R4 in parallel. The total resistance of the external circuit is 5 ohms, and the voltage at the end of the circuit is 10V.

The voltage at both ends of each resistor is 5V, and according to the shunt principle, the current through the four-bar resistor is i1 = i4 = 1 3a and i2 = i3 = 5 3a.

Then the current flowing through R3 is shunted at point D, part of it flows through R4 (1 3A), and the other part meets with I1 through a switch and flows through R2 with a magnitude of 4 3A.

1.From the electric potential energy, the electric field force from b to c does zero work. From A to B, the electric field force does negative work and the mechanical energy is lost.

2.With O as the center of the circle and Oa as the radius as the circle, we can know that in the electric field of the point charge O, the potential energy of the charge at point A increases from A to B, so the electric field force does negative work. Or it can be judged by the angle between the applied force and the direction of the path.

It's a very simple topic, you shouldn't think much about it, it's important to think for yourself.

1.The total energy is conserved, the electric potential energy increases, and the mechanical energy decreases.

The electric field of the point charge is already known, the potential energy of a-b increases, and the electric field force does negative work. The b-c potential is unchanged, and the electric field force does not do work, which is judged by the distance between the ball and the point charge O point.

Answers B and C are correct.

A is not that this is not a uniform motion, the acceleration has been changing again from the diagram can be analyzed The acceleration of the ball in CD is always to the left, so the ball decelerates first, B may be the initial velocity of the ball is very fast, to D The velocity is not reduced to 0, and then rush out of D and go towards E, here the speed between CD is required, so B is correct (the latter one does not care).

For c, the ball decelerates to the right, and the velocity decreases to 0 before d. Then the acceleration to C (acceleration to the left), so C is correct, and D, the velocity is not 0, it is impossible to accelerate towards C, hence BC

The answer is yes.

Analysis: At point d, the electric field strength of the electric field generated by (4q) and (q) is equal and opposite in direction, and the direction of the combined electric field strength in the cd segment is to the right, so the negatively charged particles are decelerating from c to d (not uniform deceleration motion, because the combined field strength is small).

1. The vertical direction of the ball is affected by gravity mg, and the electric field force fe is horizontally to the right, and the angle between the resultant force f of the two forces and the vertical direction is 37°.

Synthesis with force, yield: Fe mg = tan37° = 3 4 fe = 3 mg 4

2. The vertical direction of the ball is uniformly decelerated under the action of gravity, and the vertical velocity of the highest point is vt=0

The time taken to reach the highest point is t: vt=vo-gt; t=vo/g

The ball moves horizontally to the right and accelerates uniformly under the action of the electric field force with an acceleration of a=fe m=3g 4

Horizontal displacement in time t x=at 2=(3g4) (vo g) 2=3vo 8g

The work done by the electric field force = fe*x=(3mg4) (3vo 8g)=9mvo 32

The amount of change in electric potential energy = - work done by the electric field force = -9mvo 32 (the electric field force does positive work, and the electric potential energy decreases).

3. The vertical direction of the ball is uniformly decelerated under the action of gravity, and the ball is horizontally to the right under the action of electric field force to do a uniform acceleration movement, and the movement of the ball is the synthesis of these two movements. The relationship between the instantaneous velocity v of the synthesis and the horizontal velocity v and the vertical velocity v is:

v²=v∥²+v⊥²

v∥=3gt/4

v⊥=at=vo-gt

v²=(3gt/4)²+vo-gt)²

25g²t²/16 -2vogt +vo²

Transformed into a quadratic function to find the extremum problem, the opening is upward, and v has a minimum value.

v takes the minimum value, t=-(-2vog) 2(25g 16)=16vo 25g

During this time, the ball rises in height h=vot-gt 2=272vo 625g

Horizontal displacement of the globules x=at 2=(3g4) (16vo 25g) 2=96vo 625g

(1) Since the angle between the direction of motion and the vertical angle is 37 degrees when the static release is stationary, then tan37=qe mg

The electric field force is mgtan37=3 4mg, and the direction is horizontally to the right (2) after the time to the highest point.

t=v g, acceleration due to electric field force, a=qe, m=3, 4g, horizontal displacement of the ball s=

The amount of change in electric potential energy = the work done by the electric field force = qe * s = 3 4mg * 3v 2 8g = 9mv 2 32

3) What is done is a parabolic motion similar to an oblique toss.

When the partial velocity of the velocity in the direction of the resultant force is zero, that is, the velocity decomposes along the parallel angle of 37 degrees and the vertical angle of 37 degrees, and when the perpendicular partial velocity is equal to zero, the velocity of the ball is the smallest.

The resultant force on the ball = mg cos37 = 5 4mg total acceleration a = 5 4g

Velocity decomposition, the partial velocity along the direction of acceleration v1=vcos37 when v1=0, at=v1 t=16v 25g is perpendicular to the direction of acceleration is a uniform linear motion s=v2t=vsin37*16v 25g=48v 2 125g

Therefore, its position is at an angle of 37 degrees to the horizontal, and the upward distance is s=48v2 125g.

a is the acceleration resulting from the electric field force.

1): Analysis of the force on the ball (triangle rule).

Electric field force direction: to the right.

Size: Fe=3G4=3mg4 (Fe stands for electric field force).

The velocity of the ball when moving to the right TS: v=at=(3g 4)*(v g)=3v 4

There is a kinetic energy theorem (horizontal):

We=(mv2) 2 (we denotes the work done by the electric field force on the ball, i.e., the amount of change in electric potential energy).

Substitution data: we=(9mv2) 32 (change, no negative number).

3): Curvilinear motion, but not a flat-throwing motion. I use the derivative method to solve (interested, please ask, of course, you can also use the upstairs method), and calculate: t=16v 25g, v has a minimum value, so:

s=(1/2)at^2=96v^2/625g

Please smile (please ask if you want a picture).

Attached: Definition of Quasi-Flat Throwing Motion:

Conditions for a flat-tossed movement:

Subjected to constant force;

The direction of the initial velocity is perpendicular to the constant force

For example, the pendulum ball swings around the suspension point in the vertical plane, and the rope suddenly breaks when it reaches the lowest point, after which the pendulum ball moves; The rotation of an umbrella around a vertical axis, the motion of the water droplets on the edge of the umbrella after being thrown out (regardless of air resistance) and so on are examples of flat-like movements.

Horizontally to the right.

2.Because the ball is only subjected to gravity in the vertical direction, t=v g, and the ball accelerates evenly to the left, a=gtan37

So ep=eqd=1 2mv 2tan 2 37=9 16mv 23Flat toss-like motion The speed is minimal at the highest point. Tired. ）

Haha, it's so kind, I'll take a look.

a No, the particles in Figure B will deviate from mn when they are subjected to a force to the left horizontally.

b Yes, the particle is stressed in the pq direction. This can be achieved by selecting the appropriate initial velocity.

c Yes, doing a uniform circular motion requires a centripetal force pointing to the center of the circle. Can be satisfied.

d No, the net force of the particles in Figure B does not point to the center of the circle.