Electric Field Force 10, Force Analysis Question on Electric Field

The key to choosing the first 3 options in this question is not to be fooled by the condition that "the magnitude of its acceleration is magnitude." Because it only requires the change of energy, and what electric potential energy, overcoming the electric field force to do work, etc., you just need to know that the uniform electric field is a conservative field, and the change of potential energy is only related to the beginning and end positions, so as long as it moves the s distance, the change in electric potential energy is qes.

As for the change in kinetic energy, it should be easy to calculate according to Niu Er. The three equations can be:

vt 2=s, kinetic energy=mvv 2 (three unknown quantities v, t, kinetic energy) The process of solving the equation does not need to be written, right?

If the mass of the object is m, then according to the acceleration, the combined external force of the object can be known.

Since the relationship between the direction of motion of the object and the direction of gravity is not known, the electric field force does negative work, and the magnitude is EQS, then the electric potential energy increases EQS

According to the kinetic energy theorem, the kinetic energy decreases when the external force is used to do the work.

Choose C, D

On the dotted ring, take a charge element dq= dl, and the electric potential generated by the charge element at point A is d = (1 4 0) r + oa = (1 4 0) ( dl 2r) at the root number of dl. The potential of point a obtained by integrating the two sides is: a=(1 4 0) l 2r=(1 4 0)( 2 r 2r)= 4 0.

In the same way, the electric potential at point b can be obtained: b= 6 0.

The work done by the electric field force of the charged particle q moving from a to b is: w = q ( a - b) = q ( 4 0 - 6 0) = q 12 0.

Because the energy of the electric field is active, the kinetic energy is generated by the potential energy. Therefore, the potential can be smaller. Note: The potential can be smaller, but the force is larger.

The algebraic sum of the potentials of all charges on the ring at points A and B can be calculated first, and then calculated with w=qu.

In order for the ball to pass through the tube without collision, it means that when the ball is above the nozzle, the horizontal velocity of the ball must be 0.

Pushing back from the solution goal, the solution idea is probably.

The initial velocity v0 (which can be obtained by the formula of flat throwing motion) (known) l, t h

The kinetic energy sought ek

Electric field strength e (Newton's second law) a (motion formula) v0

Such an arrow represents an equation, and the column equation will do:

Solution: by the kinetic energy theorem ek=mv0 2+mgh-eql

by Newton's second law eq=ma

By the motion formula v0-at=0

v0t=l ④

gt 2 = h 2 (in the equation t represents the time from the beginning of the motion of the ball to the nozzle).

Solved by the equation ek=mgh

The answer is right.

If you don't understand, you can ask, hehe

When landing, the horizontal velocity is zero, while the vertical direction only gravitational work is done, so the answer is correct.

Because the oil droplets are negatively charged, the electric field force they receive is positive on the x-axis! Therefore: (1) The oil droplets shot out in the positive direction of the x-axis will move in a straight line with uniform acceleration, and the electric field force f=eq, a=f m, s=v0t+1 2at 2, and the time can be found by solving this equation!

2) Because the direction of force is fixed, the kinetic energy must be the largest in order to have the greatest displacement, so the kinetic energy of the oil droplets shot out along the direction of the force is the largest! It's the case of 1 question, and the time of 1 question is out! So in the end, the velocity can also be found, just calculate the kinetic energy!

Decompose v, vertical direction: v1=v cos@=2gh under the root, so the high defeat key degree h is obtained

Horizontal: v2=v0+at=v sin@, and t=2h under the root is better than g, a=eq m, so the answer can be found e.

Only support the addition of the rearmost of the 2nd floor, the rest is a mess of ......Note that the title says: the particle moves from A to B, and the trajectory is the dotted line in the figure!

The fundamental problem is that the title doesn't say that particles move from a! The trajectory in the diagram may be part of its trajectory, and the particle may:

1.Pass A from elsewhere and then move along the trajectory past B

2.Start the movement from A and then pass through B

What happened before A is not mentioned in the title, so we don't know the initial velocity at all.

We know that the concept of initial velocity is the velocity that the particle has when it starts to move, and the instantaneous velocity direction of the particle moving in a curve is in the tangential direction, so the velocity of the particle must be in the tangential direction at point a, that is, the particle has velocity at a! But the important thing is that the title doesn't tell you the ...... that particles move from a

To take a simple particle, if you start walking on the road with a bag of rice on your back from a standstill, and the rice bag is cut and the rice starts to fall out, then the place where the rice bag is cut is recorded as point A, and then you continue to walk for a while and then it is recorded as point B, and when you look back, the trajectory of the rice grains is only from point A, but you start moving from stationary ......So the initial velocity can be 0.

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.

Select C to analyze the force of the balls separately.

tana=kq1q2/m1gr*2

tanb=kq1q2/m2gr*2

Because of AM2

c The electric field force is the same, that is, the force on each ball is the same in the horizontal direction.

The ball is affected by three balancing forces, so the small vertical angle is subject to gravity.

Force analysis is sufficient.

From the inscription, it can be seen that pn is on the same equipotential surface, from n to p the electric field force does not do work, only gravity does work, and mn is smooth, let the distance between dm be x, then the work done by the electric field force from d to m is qex=2mgr, and qe=mg 2, x=4r can be obtained. Similarly, if the ball is released from the right side of m, 5r, since the maximum static ground grip friction is equal to the sliding friction force, the small ring will eventually rest on pq, and the work done by the electric field force is equal to the work done to overcome gravity, the work done to overcome friction, and the work done to overcome the electric field force, then 5qer=2mgr+ mgs+qes, and qe=mg 2, then the distance s of the movement of the small ring on pq can be obtained, so that the work done by friction can be obtained.

2) Release the guess point to m, and the electric field force does positive work w1=mg 2*5r;

The gravitational potential energy from point m to point p does negative work w2=mg*2r; The electric field force first does positive work and then negative work, which is zero in total;

The kinetic energy at the point of p is ek=mg 2*5r-mg*2r=;

From the point p to the stationary point, the distance is L, the electric field force does negative work w3=mg 2*l, and the friction force does negative work w4=u*mg*l;

According to the kinetic energy theorem, -mg 2*l-u*mg*l=;

Final state: The travel force of the electric field cavity is equal to the maximum friction force of the static Mo Wu Zhao stool, mg 2=u*mg. then, u=;

Substituting the above equation, it can be found, l=r 2;

Therefore, the work done to overcome friction in this process is w4=u*mg*l=mgr 4.