Electric field force work problem!! Here comes the master!!

This formula is very vaguely written. In short, the work done by the electric field force can be understood by analogy with the work done by gravity, for example, gravity does positive work, and the gravitational potential energy decreases; Gravity does negative work, and the gravitational potential energy increases. In the electric field, there are positive and negative charges, for the positive charge, the electric field force does positive work, and the electric potential energy decreases; The electric field force does negative work, and the potential energy increases.

For negative charges it is the opposite.

If in a simple electric field, only the electric field force does the work, then the particle's electric potential energy and kinetic energy are converted into each other, for example, when a positive charge exists only with the electric field force, the electric field force does work on it, and it has to move, so the electric potential energy decreases and the kinetic energy increases.

If only gravity and electric field do the work in the composite field of the electric field and the gravitational field, the work done by the gravitational force and the electric field force can be analyzed one by one as above, so that the kinetic energy, gravitational potential energy, and electric potential energy of the particles increase or decrease.

You should be new to physics, in fact, this is not difficult. To the left of the equal sign is the work done by the gravitational force and the electric field force, and the corresponding force doing the work must cause the corresponding energy change. The calculation of kinetic energy depends on whether there is a change in the initial and ending velocities, and the initial kinetic energy is subtracted from the final kinetic energy.

When doing this kind of problem, don't forget that there are other conservation laws of energy, and there are just a few formulas. Don't be afraid, it's not hard.

The second w should be the work done by the electric field force, right? It is enough to understand the law of mechanical energy and equilibrium.

All work processes are energy conservation, just a transformation process.

The solution to this kind of problem of conservation of energy is very simple, analyze the force, see which forces do the work, and then write the equation, just pay attention to the plus and minus signs.

ub = 0，ud = q/(4πε0 3r) -q/(4πε0 r) = - q/(6πε0 r)

The work done by the electric field force.

a = 1*(ub - ud) = q/(6πε0 r)

When elastic metal pellets A and B are not in contact, the Coulomb force f1 = (kq1q2) r = (9kq) r

After the contact between elastic metal balls A and B, the power is first neutralized and divided into equal parts, so the charge of the two balls is 4q, which is mutually exclusive.

After the contact between elastic metal pellets A and B, the Coulomb force f2=k=(q1q2) r =(16kq) r

From the acceleration a=f m, we get a2:a1=16 9

The two balls are close to each other because of mutual attraction, and the electric charge of the two balls after contact is 4q, and the electric properties are the same and mutually repellent, and the instantaneous acceleration is (9q2 r2) m when released, and the instantaneous acceleration is (16q2 r2) m, so it is 16 9 times.

Do a linear motion, the net force is 0, directly let the direction of the electric field upward, and the generated electric field force and gravity cancel out.

f=mgeq=mg

e=mg/q

Because it is a positive charge, the direction of e is upward, so that the direction of the electric field force can be upward.

In a vacuum, the magnitude of the interaction force between the two stationary point charges Q1 and Q2 is proportional to the product of Q1Q2 and inversely proportional to the square of the distance r between them. Among them, it is important to note the conditions for Coulomb's law to be true: in a vacuum, it must be a point charge.

Coulomb force formula f1=kq1q2 r 2 k= where q1 and q2 are the individual charged charges of the two objects, r is the distance between the two objects, and k is a constant.

This problem is f=k qa qb r 2 ,qa=2qb so there is f=k 2qb qb r 2=2 k qb 2 r 2, bring in the number, qb 2

Boost power: p=w t=490000 140=3500 translation power 0

mgh=pt

t=mgh/p

150s

One thing you have to understand to do this kind of problem is that the force and the direction of motion must be on the same straight line!

For example, the first question, the crane lifts the goods, the tension of the rope and the gravity of the goods are in a straight line, so we calculate the work done by the crane, as long as the force in the vertical direction is multiplied by the distance moved in the vertical direction, the distance of horizontal movement does not need to be calculated! w=mgh=5000*10*10=490000j

Lifting power: p=w t=490000 140=3500

The premise of a uniform magnetic field between charged parallel plates is that the parallel plates are infinitely large, that is, the distance between the plates is much smaller than the size of the plates. Here are two thin rods, then you can look at the thin rods as countless charged points, and then it is easier to analyze, the field strength a>b>c, you can also use the integral to calculate. As for the electric potential, the electric field lines are perpendicular to the AC wires, so the electric potential is equal.

The dotted line is actually an equipotential line, the top view of the electric field distribution formed by the thin rod is the same as the plane diagram formed by the different charges, as for the metal plate, I think you must know, why the answer is A, I think it is B, let me think about it again.

Analysis: The particle is moving in a curvilinear motion, and the direction of the electric field force (resultant force) it is directed (roughly) towards the inner side of the curve (i.e., the concave side).

Corresponding to the situation in the diagram, the direction of the electric field force experienced by the particle is downward, so the particle is positively charged.

The density of the electric field line indicates the magnitude of the electric field strength, obviously the electric field strength at point B is greater than the electric field strength at point A, then the force of the particle at point B is greater, that is, the acceleration of the particle at point B is larger.

Since the force experienced by the particle is downward, and the particle moves from A to B (the tangent of the trajectory of the dashed line in the direction of velocity), it is easy to see that the electric field force does negative work on the particle, and the velocity decreases, so the velocity of the particle at point A is larger.

If the particle moves from A to B, the electric field force must follow the direction of the electric field line, because the combined external force of the object moving in a curve points to the direction of the bowstring, and it cannot point to the direction of the back of the bow, so the particle must be positively charged; The electric field at point B is dense and the field is strong, the electric field force is large, and the acceleration is large, so the acceleration at point B is larger; From A to B, the electric field force does negative work, and the particle velocity at point A is larger.

positive charge; The acceleration at point b is larger; The velocity at point A is larger.

Select B for this question. In this case, the charged ball can move in a straight line, the ball receives the attraction from the left side to meet the situation, the ball is balanced by the electric field force to the left and gravity, the direction of the resultant force is opposite to the velocity of the ball, the velocity of the ball decreases, the kinetic energy decreases, the ball rises, the gravitational potential energy increases, the electric field force does negative work, the electric potential energy increases, the kinetic energy is converted into electric potential energy, the velocity becomes smaller, and the mechanical energy decreases.

Gravity does negative work, and the gravitational potential energy increases; The electric field force is positive, and the electric potential energy is reduced; The kinetic energy of the horizontal velocity increases, and the kinetic energy of the vertical direction decreases, depending on the amount of charge and the weight of the ball.