High School Physics Potential Questions, Urgent Answer Process, Thanks

The electric field force is negative to the 3rd power J, which means that the electric potential energy of the charge at point b at 2cm of a is +, then the potential energy after release), and from the zero potential to the 3000V potential to the 3rd power j of the negative work, this shows that the 3rd power j of uq=, calculate q = +6 * 10 (-7)c, and at 2cm the acceleration magnitude is 9 10 to the 9th power m s2, you can get kqq r 2=am, the mass of b charge is m =, according to the calculation of the maximum velocity is the 3rd power m s

At 2 cm, the electric quantity q=w u is calculated as q, the force f is calculated by Coulomb's law f=kqq r2, and then the mass m of charge b is calculated from f=ma, and then the electric field force does work w=1 2mv2, and v (w is equal to the 3rd power).

The formula for electric potential energy is: ep=q. (Did the formula teacher talk about it?) EP denotes electric potential energy, Q denotes electric electricity, denotes electric potential. Note that q in the formula has positive and negative values.

There is a little p in the electric field formed by the negative charge q, and the potential along the direction of the electric field line is reduced, so for the negative charge, the potential close to it is relatively low. Then the infinity is 0, and the near negative charge is lower, then it should be negative. So the electric potential at point p is negative.

is a negative value, if you put a positive charge, q is a positive value, one positive and one negative, then EP is a negative value.

is a negative value, if you put a negative charge, q is a negative value, two are negative, negative negative is positive, then EP is calculated as a positive value.

If the charged properties of the two point charges are opposite, then the direction of the electric field force does not change regardless of whether the two point charges are placed on Ab or Cd, and it is impossible for the particle to reciprocate.

If both point charges are positively charged, the two point charges should be on ab; If both point charges are negatively charged, the two point charges should be on the CD. The direction of the restoring force is always opposite to the displacement. However, unlike the simple harmonic oscillator, the magnitude of the restoring force is not proportional to the magnitude of the displacement. As shown in Fig.

First of all, the two point charges cannot be opposite in nature, otherwise the particles without muzzle velocity positively charged will inevitably move to the negative point charge, and will not make a round-trip in the middle, analyze cd, if you want to make a round-trip motion, you must be forced, the direction points to the equilibrium position, that is, the o point, the analysis of its motion, is to accelerate to o first, reverse force through o, decelerate to b to stop, and then reverse acceleration to o, so that the cycle can do round-trip motion, the key is with the force, c is the repulsion force, d is the gravitational force, all meet the requirements, about the electric potential, Here the system is the conversion of kinetic energy and electric potential energy, which is conserved by energy, and the kinetic energy at ab is 0, so the potential energy must be the same, so CD is chosen

Typing is not easy.

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To put it bluntly, that is.

The electric field force does positive work, and the potential energy of the charge decreases, and the electric field force does negative work, and the potential energy of the charge increases.

For the electric field formed by the positive charge, the positive tentative charge is close to the field source, because the positive charge is close under the action of repulsion, so the electric field repulsion does negative work, and the electric potential energy increases. The positive test charge is far away from the field source, because the positive charge is far away under the action of repulsion, so the electric field repulsion does positive work, and the electric potential energy decreases; The negative tentative charge is close to the field source, and the positive charge is close under the action of attraction, so the electric field attracts positive work and the potential energy decreases. The negative tentative charge is far away from the field source, because the positive charge is far away under the action of attraction, so the electric field attraction does negative work, and the electric potential energy increases;

For the electric field formed by the negative charge, the positive tentative charge is close to the field source, and the negative charge is close to the force of attraction, so the electric field attraction does positive work and the potential energy decreases. The positive tentative charge is far away from the field source, because the negative charge is far away under the action of attraction, so the electric field attraction does negative work, and the electric potential energy increases; The negative tentative charge is close to the field source, because the negative charge is close under the action of repulsion, so the electric field attracts negative work, and the electric potential energy increases. The negative tentative charge is far away from the field source, because the negative charge is far away under the action of repulsion, so the electric field repulsion does positive work, and the electric potential energy decreases.

As for the electric potential, it is better to judge, the closer you are to the positive charge, the higher the potential, and the closer you are to the negative charge, the lower the potential.

If there is a point of charge that does positive work during the process of moving from a charge close to the field source to a charge away from the field source, the electric potential energy decreases.

Regardless of positive or negative charge, as long as the electric field force does positive work, the electric potential energy decreases.

If it is a negative charge that moves from a charge close to the field source to a charge far away from the field source, the external force does positive work, and the potential energy between the charges increases. It's like when you hold up a heavy ball, and the relationship between the earth and the heavy ball, you're doing the right thing.

First of all, A's option is wrong. The proof is as follows: the field strength is perpendicular to the vertical plane and the electric field force is also perpendicular to the vertical plane in the electric field line distribution diagram of two equally dissimilar charges.

That is, that is, if the charge at that point moves on this perpendicular plane, no matter where it moves from there, the electric field force does not do work because the electric field force is always perpendicular to this perpendicular plane. The electric field force does not do work, and the electric potential energy does not change. Therefore, as long as you take two points on the middle vertical line, which can not only meet the topic, but also easily prove the wrong statement that the field strength must be the same, you will be proved if you deliberately take two points that are not the same.

Secondly, B's option is also wrong. The proof is as follows: the charge at this point can move from point A to other points first, so that the electric field force can do positive and negative work at will, and then move from that other point to point B, so that it can not only explain the statement of C, the statement that the electric field force is always perpendicular to its direction of movement, but also that the work done by the electric field force will cancel each other, that is, the electric field force is equal to no work, so there is naturally no change in electric potential energy.

Therefore, it is also proven.

The electric potential energy does not change, indicating that the total work done by the electric field force when the charge moves in the electric field is zero. This is the basis for judging the problem, and you can deduce from this basis Of course, there is a prerequisite for solving this problem, which is that you must understand the various types of electric fields. If you need a more detailed explanation, you can find me at I am an online physics tutor -57589-232

A is false, the electric potential has nothing to do with the electric field strength, along the direction of the electric field line, the electric potential gradually decreases. The strength of the electric field is only related to the magnitude of the electric field force. So a mistake.

C false, ditto, it can move at will, in any direction, as long as it can return to the equipotential plane at the end. But the direction of the electric field force is constant, and the direction of the electric field force experienced at each point is fixed. Therefore, it is wrong to say that the direction of the electric field force is always perpendicular to the direction of motion.

In fact, you can compare the electric potential to the level of water, as if the current is a flow of water, and the electric potential decreases when it decreases in the direction of the current.

The first sentence is wrong. It is also wrong that the second sentence does not presuppose the first sentence.

Synthesize a sentence:

Following the direction of the current, the resistance potential is encountered to reduce the IR. That's right.

Resistance can only consume electrical energy, so the potential of the current passing through the resistance decreases.

The power supply is different, it can be a rise or a drop in potential

The key to this problem is to clarify the conservation of energy in the process of electric potential energy and kinetic energy. The time is the same, which is the relationship between the x-axis and the y-axis of the two sub-motions bridge. So, study the process of particles from p to c'.

X-axis direction v*t=2cm

The y-axis direction 1 2*a*t*t=1cm where a=eq m is solved to t=m*v e*q and a*t=v

So c'The point x-axis and y-axis score velocities are equal.

So ekx=eky=ek

then c'The kinetic energy of the point is 2ek b.

Because the space is a uniform electric field, then aa'The drop in potential energy to bb' is equal to that of bb' to cc'.

So by bb'The kinetic energy of the time is therefore awrong.

Because of BB'is a zero-potential surface so aa'The face is the face, so C is also wrong.

The same goes for CC'is the correct option.

As mentioned in the above process, the potential sign of the particle changes when passing through the zero potential energy surface, but the electric potential has no direction, it is a scalar quantity.

The positive or negative work should be judged from whether the direction of motion is the same as the direction of the force, and negative work should be done when the motion is obstructed. The initial potential energy is subtracted from the end potential energy during the operation, and negative work is done when the result is negative.

Emphasis: When the electric potential energy decreases, the kinetic energy of the particle increases, which is the essence of doing positive work. So in acceleration.

Because the particle is positively charged, there is an electric field direction A pointing to C, so the electric potential of A is high (>0), and C < 0; The electric potential (scalar) has no direction, only size, and the electric potential keeps getting smaller, so the electric potential energy decreases (the electric field does positive work), so it keeps accelerating.

Because Pa'=A'C', according to the formula of accelerated motion, we can know that the horizontal and vertical velocity at point C' are equal (but it cannot be said that the horizontal and vertical kinetic energy are equal, and the kinetic energy is also a scalar quantity), so the kinetic energy is 2ek, and the increased kinetic energy is converted from the electric potential energy, and Ab and BC are converted respectively, so the electric potential energy of C' is 0).

The field strength is a physical quantity that represents the nature of the electric field force, which reflects the strength of the electric field and the directional electric potential represents the nature of the electric field energy, and the electric potential difference reflects the electric potential of each point on the potential surface of the electric field's ability to do work, but the field strength is generally different.

In the electric field of an isolated point charge, the equipotential surface is a series of concentric spheres centered on the point charge, and the points on the sphere are equidistant from the point charge, so the electric field strength is also the same.

The field strength is to see the density of the electric field line, the sparse weak, the dense strong, and the equipotential surface is the equal electric potential, which is not necessarily related.

There's no direct relationship between electric potential and field strength, so there's a lot of that.

: acceleration increases the beam plus the draft, so a is wrong, b: the charge is positive and negative is uncertain, so b is wrong, c: the kinetic energy of accelerating the side bending is increased, the electric potential energy is reduced, so c is wrong, d: the kinetic energy is increased, and the positive work is done, so d is correct.