High School Physics Electrostatic Phenomena Questions, High School Physics Problems Application of E

1.The outer surface is positively charged, and the inner surface is not charged.

After the negatively charged ball is inserted, due to electrostatic induction, it is the inner surface of the metal spherical shell that induces the opposite charge (i.e., positive charge), and at the same time, the outer surface is negatively charged at this time due to the conservation of charge. When touched by hand, the negative charge on the outer surface is directed to the ground by the hand, while the positive charge on the inner surface is not directed away due to the attraction of negatively charged balls. When the pellet is removed, the positive charge from the inner surface is distributed to the outer surface uniformly because the surface area of the outer surface is greater than the surface area of the inner surface.

2.The outer surface is negatively charged, and the inner surface is not charged.

The first step is the same. After the negatively charged ball comes into contact with the spherical shell, part of the negative charge on the ball is transmitted to the spherical shell, so that the spherical shell is negatively charged (because it is originally electrically neutral, the charge charge on the inner and outer surfaces is the same, and the electrical properties are opposite), and the charge distribution on the spherical shell is not uniform. The third part is the same.

If you don't understand, you can hi me.

If it is not in contact, due to electrostatic induction, the outer surface of the metal ball is negatively charged, and the negative charge is taken away when the hand is in contact, and after the charged ball is removed, the metal ball is positively charged, and the outer surface is positively charged and the inner surface is not charged under the action of its own electrostatic induction.

If it is contacted, it forms a whole, due to its own electrostatic induction, the electrostatic charge is concentrated on the outer surface, and after being guided away by the hand, it is not charged, so the inner and outer surfaces are not charged after the charged ball is removed.

The outer surface is positively charged, and the inner surface is not charged.

The outer surface is negatively charged, and the inner surface is not charged.

This is called electrostatic shielding, and the charge is only distributed on the outer surface.

The outer surface is positively charged, and the inner surface is not charged ......

If it comes into contact with it, it will not be charged ......

The electroscope does not respond, how can there be a reaction when there is no electricity?

The answer is to choose C No, after the outer wall contact, E has a positive charge and contacts on the inner wall of A. 1. The charged body is electrified, because it is the same kind of charge, so it repels each other, and is finally distributed on the surface of the charged body and is in a stable state. 2. After 1 contact with A, similarly, A is charged on the outer wall, so the contact on the inner wall, because the same charge is in contact, what is obtained is only the mutual transfer between the same charge, and the transfer amount of charge depends on the amount of band point of the conductor carrying the charge and the external geometry.

3. When E is in contact with A, the positive charge is in the inner wall of A, and then transferred to the outer wall of A, so the amount of charge on the inner wall is always zero at this time, and the inner wall of A is always free of charge. 4, so in the process of transfer, the inner wall of A in the two charged bodies, which is always zero, can obtain the maximum charge transfer. To sum up, choose C

That's the question, hehe, that's it.

First of all, it is clear that the conductor rod and the ground are insulated, so although there is an induced charge on the conductor rod before the finger touches, the net charge is 0, this is no problem, right?

Secondly, the hand touches it, no matter whether it touches **, it is to connect the conductor rod with the earth, so after touching, the contact part becomes the far ground end (the same is true for touching a), so after the hand touches it, the near ground end will induce a negative charge, and the far ground end has no charge because it is connected to the earth, and the net charge on the conductor rod is not 0, but negative.

After removing the hand, it is equivalent to insulating the conductor rod again, so the negative charge on the conductor rod has nowhere to go, and it will be distributed on the surface of the conductor rod, so the negative charge is a thing.

Do you understand?

The finger is one with the person. The fingers are in contact with the conductor rod AB, and the person stands on the earth, which forms a whole (AB + person + earth). Due to the proximity of the whole to q, the free electrons in the whole move to the end of the whole that is close to a positive charge (i.e., the conductor rod AB), resulting in the AB terminal being negatively charged.

The same is true for the two blanks, in fact, the questioner divided the two blanks in order to confuse the students.

After the conductor is close to the positively charged Q, the phenomenon of electrostatic induction will occur, the left end of AB, that is, A, will be negatively charged, because the original is not charged, so B will be positively charged, after touching B with your hand, the positive charge on B will move to the hand, after removing Q, AB will be charged depending on the A end, so it is a negative charge.

The same can be analyzed to get the second answer.

Experiments have measured that the conditions for lightning to occur are that the electric field strength exceeds v m and the height above the ground is 300 m, so for lightning to occur in this area, the potential difference is u = x 300 = v

1. If object A and object C repel each other, object B must be able to (attract) object C, and object B must have a (positive) charge.

2. In summer, the friction between two clouds causes them to carry an electric charge, one of the clouds has a negative point due to the acquisition of electrons, and the charge carried by it and the charge carried by the other cloud will attract each other.

Same-sex electric phase repulsion, opposite-sex electrical phase attraction...

1. A and negative charge repulsion Description A is a negative charge then B positive charge and C attract.

2. Because electrons are conserved, one positive and one negative attract.

This is a problem of three charge balances, which follows: the large clamp is small, and the same clamp is different (the large amount of charge is on both sides, and the same kind of charge is on both sides), so the C charge is positive, and the charge of C is Q, and the distance R from the B charge is R, then the gravitational force of B on C is equal to the repulsion of A to C, i.e., (KQQ) (R 2)=(4KQQ) [R+L) 2], the gravitational force of C on B is equal to the gravitational force of A on B, i.e., (KQQ) (R 2)=(4KQQ) (L 2), and the solution of the two equations gives R=L, Q=4Q, Therefore, the net charge of the electric limb of c is +4q, which is l away from b

After analysis, it can be seen that to achieve the C charge, it must be located on the right side of the -q shirt. Let the C distance A be R1, let the C distance B be R2, 4KQQ (R1)2=KQQ (R2)2, R1+R2=L. It can be counted as a silver outbreak.

The first point charge c should be in the same line as a and b, otherwise it would be impossible for any of the charges to be in equilibrium by two non-zero non-collinear forces. If you analyze it carefully, you will find that if C is negatively charged, then A is between B and C, but at this time, Ca sets B, C spacing is X, then A,C spacing is X+LLet the charge of c be q again. There are according to the equilibrium conditions.

The answer is to solve 1 4t time from a quarter of a period. We know that at -x0 the velocity of the ball is 0, and then under the action of the electric field force, the acceleration is a=f m=qe m=q 0 md (equation) uniform acceleration linear motion reaches x=0, so there is 1 2a* =x0 so the time to solve t is 1 4t, and t can be solved by bringing in the x0 solved earlier.

If you don't understand it, don't think about it, the finale question is not for everyone, and if you understand one, it will basically not help the next finale question.

Newton's second law, because the field strength is constant, he does a linear motion with uniform variable speed.

In a conductor that achieves "electrostatic equilibrium", the internal field strength of the conductor is zero, and the charge is only distributed on the outer surface of the conductor, and the sharper the point, the greater the charge density.

The function of the wire here is to make the ball and the spherical shell become a whole, so the ball is equivalent to the outer surface of the conductor when it is outside the spherical shell, so it is positively charged.

High school physics, applied knowledge of electrostatic phenomena.

1 Electrostatic equilibrium.

The state in which the free electrons in the conductor no longer move in a directional manner is called the electrostatic equilibrium state

2 Electrostatic balance characteristics.

1) A conductor in electrostatic equilibrium with zero internal electric field strength;

2) The whole conductor in electrostatic equilibrium is an equipotential body, and its surface is an equipotential surface

3 Charge distribution characteristics of electrostatically balanced conductors.

1) The inner and outer surfaces are differently distributed, the charges are distributed on the outer surface, and there is no charge inside the conductor

2) Affected by the shape of the conductor, the charge density is high in the non-smooth area On the surface of the conductor, the sharper the position, the greater the density of the charge, and the concave position has almost no charge

3) The "far and near" charges are opposite in their electrical properties and have equal charges

As shown in the diagram, when two or more conductors that were originally insulated from each other are contacted or connected by wires, they can be regarded as one large conductor, and the "contact of a conductor by hand" mentioned in some questions is actually the conductor forming a large conductor through the human body and the earth

4 Tip discharge.

1) Ionization phenomenon: the phenomenon of separation of positive and negative charges in the molecules of gas molecules in the air under the high-speed impact of charged particles

2) Tip discharge: The charge at the tip of the conductor attracts the charge in the air that is opposite to its symbol, and the charge on the tip is neutralized and loses its charge, a phenomenon called tip discharge

5 Electrostatic shielding.

A cavity conductor in an electric field has no electric field in the cavity because there is no electric charge on the inner surface, and this phenomenon is called electrostatic shielding

There are two cases of electrostatic shielding:

1) The external electric field of the conductor does not affect the inside of the conductor, that is, the superposition of the external electric field and the induced charge on the surface of the conductor at any point in the conductor is zero

2) The internal electric field of the grounded conductor does not affect the outside of the conductor because the induced charge outside the conductor is repelled into the ground, so that the superposition of the induced charge and the induced dissimilar charge on the inner wall at any point of the electric field strength outside the conductor is zero