Ask for a physics problem on electrostatic shielding, and give 100 points if you are good

First of all, I would like to show respect for the landlord's attitude towards learning.

For your question: 1 The cavity is grounded, and the earth is generally taken as a reference, and the electric potential of the cavity is zero, so that the potential difference between any two points in the cavity is 0, and the electric potential is also 0

2 If you think about it this way, what happens if there is an electric charge on the inner surface and it's not zero, and an electric potential is generated, so there is no free charge on the inner surface and in the cavity.

3 If it is not grounded, after electrostatic equilibrium, there is a free charge on the surface of the cavity, and the electric field formed by these charges is in the opposite direction to the electric field formed by the charged body q in the cavity, and the magnitude is equal, so there is no free moving charge in the cavity.

1.Grounding means that the cavity conductor and the earth form a large conductor, and their equal potentials are 02You make a common mistake that grounding does not mean that the charge will necessarily run down the ground line to Earth. The right way to think about it should be:

Think of the cavity conductor and the earth as one large conductor, and the whole conductor is subjected to electrostatic induction, which carries a different charge at the closest place to the charged body q, and the farthest place (on the other side of the earth) has the same charge. That is, the principle of "near and far from the same".

3.Also according to the principle of "near and far from the same", the nearest place to the charged body q of the cavity conductor has a different charge, and the farthest place from the charged body q has the same kind of charge.

1.The earth potential is 0 (this is common knowledge and must be known), and the cavity conductor is grounded, then it should have the same potential as the earth, so the electric potential energy is 0

2.Because the charge q is close to the cavity conductor, and the cavity conductor is connected to the earth, then the cavity conductor must have electrons flowing into the earth in order to maintain 0 potential (or there are electrons flowing from the earth to the cavity conductor, depending on whether the charge q is positively charged or negative), so that the algebraic sum of the charge of the cavity conductor is not equal to 0, and the cavity conductor is dotted.

Because it is a charged cavity conductor (note that it is a conductor), the internal free electrons can move freely, and the same charge repels each other, so the charge is expressed on the outer surface due to this repulsive force.

3.The cavity conductor is not grounded, the charge is also distributed on the outer surface, the side near q is electrically opposite to q, the side away from q has the same electrical properties as q, the sum of the total charge algebra is 0, and the potential is not equal to 0 (if q is negatively charged, the cavity conductor potential is negative, if positive, it is the opposite).

1.The electric potential is zero because of grounding.

2.Different charges attract each other, the charged body Q attracts the outer surface to make it charged, and the charge on the inner surface flows to the earth through the outer surface, because the conductor is an equipotential body, if the charge is inside when we draw the electric field line, it has to pass through the conductor, then the conductor does not have a potential difference? contradictory to each other.

3. If it is not grounded, the conductor has the same amount of different charge, and the quantity is equal to q. Exterior with Q heterogene.

a.The inner side of the spherical shell is negatively charged, and the outer side is positively charged.

The electric potential inside the spherical shell is high and the external electric potential is low (as can be seen from the electric field lines at this time, the electric field lines point from the positive charge to the inner side of the spherical shell, and then from the outside of the spherical shell to infinity).

b mistakenly takes the ground potential as zero, then the electric potential outside and inside the spherical shell is greater than zero. The spherical shell is an equipotential body and the electric potential is also positive, which is greater than zero. C false.

After electrostatic equilibrium, the internal field strength of the metal conductor is zero, but there are electric field lines in the internal space of the spherical shell, and the field strength is not zero. D false.

The correct answers should be A and D.

After the conductor reaches electrostatic equilibrium, the net charge will only be distributed on the outer surface of the conductor, and the inside of the conductor is the neutralization of positive and negative charges. A correct.

The field strength inside the electrostatically balanced conductor is zero, so d is also correct.

One more option, Ignore him.

Option C is correct.

If I'm not mistaken, point A is located on the outside of the spherical shell, not on the outer surface!

First of all, there is an electric field at point a, that is, the field strength is not 0, but because point A is not on the surface of the ball, the electric potentials of A and B are definitely not equal, and option A is wrong;

In the metal spherical shell entity that has reached electrostatic equilibrium, the field strength is 0 everywhere, and there is no other charged body in the cavity of the spherical shell, so there is no electric field in the cavity, that is, the field strength in the cavity is also 0, and option B is wrong;

From the previous analysis, we know that option c is correct;

Because there is no other charged body in the cavity, after electrostatic equilibrium, the charge is only distributed on the outer surface, and option D is wrong.

Due to the electrostatic equilibrium spherical shell solid and the spherical shell cavity, the field strength is 0

The combined electric field due to the applied electric field and the charge on the surface of the spherical shell is 0 within the spherical shell, both inside the solid and in the cavity

Firstly, the electrostatic equilibrium spherical shell entity neutralizes the cavity of the spherical shell, the field strength is 0 everywhere, and the potential is equal, let a positive charge be placed outside the hollow metal spherical shell as q, from the center of the sphere r, the spherical center potential = kq r = point c potential = point b potential, but the potential of a is not determined, for example, a is at infinity, the electric potential is 0, at q

Very close, the electric potential is infinite;

Regarding the field strength, BC is 0 and A is equally uncertain, but certainly proves it, since everything is caused by +Q.

For spherical shells, the induced charge is always on the inner surface. Choose C

In the 25th minute and 49 seconds, the elaboration on this issue began.

Use a grounded cavity conductor to isolate the influence of electrostatic fields inside and outside the cavity, which is called full shielding in electrostatic shielding.

If there is a charged body in the cavity conductor, when the electrostatic equilibrium, its inner surface will produce an induced charge equal to the opposite sign, while the outer surface will produce an equal amount of induced charge of the same sign. If the shell is not grounded, the outer surface will have the same amount of induced charge, at this time the electric field of the induced charge will affect the outside world, the cavity conductor can only prevent the influence of the external electric field on its interior, but can not prevent the influence of its internal charged body on the outside world, so it is called the outer shield, if the shell is grounded, even if there is a charged body inside, then the algebraic sum of the induced charge on the inner surface and the charge carried by the charged body is zero, and the induced charge generated on the outer surface will flow into the earth from the grounding, and the outside world can not affect the inside of the shell. The external influence of the internal charged body is also eliminated, and this kind of shielding is called full shielding.

It's easy to understand by drawing the narrative yourself, and the key to understanding this narrative is:

The first is that the induced charges are distributed on the inner and outer surfaces respectively, and the grounding action only affects the outer surface, and the attraction of the internal charged body has no effect on the inner surface.

The second is that the total effect of the induced charge generated on the inner surface outside the cavity is exactly offset by the effect of the charged body inside the cavity.

Note: The first paragraph is quoted from.

Well, I'll admit that this paragraph is a direct copy-paste, but what he said is already so good that I can't change it.

In the 25th minute and 49 seconds, the elaboration on this issue began.

1. Use a metal shell or metal mesh to enclose an area, so that the area is no longer affected by the external electric field, this phenomenon is called electrostatic shielding.

2. Reason: The charge of the charged object in the metal shell causes the charge of the same size and opposite properties to accumulate on the inside of the shell, and the charge with the same nature is gathered outside the shell, for the metal shell, only the charge distribution changes, and the net charge is still 0. Therefore, the charge of the charged body in the shell only has an effect within the scope of the shell, and for the external environment, its electricity is equivalent to being shielded.

Let the open hollow metal spherical shell A be positively charged, its electric potential is U, after the metal ball of electroscope A is connected with the inside of A with a wire, the electric potential of A will be forced to be equal to A, at first A is uncharged, the electric potential is 0, so a part of the positive charge on the surface of A runs to A to help A increase the potential, so A is electrified, and the platinum sheet is opened. (You can understand that the metal ball of electroscope A is connected with the inside of A with a wire, and A becomes a part of the surface of A, so it is electrified).

And B is not charged after touching the inner wall of A (because the electric potential is U after B enters the inside of A), and after taking it out, it is naturally not attached to A (of course, the electric potential will drop after coming out of A), so B's platinum sheet does not open.

Another: The electric potential of an object is the superposition of the electric potential generated by its own charge and the electric potential generated by the surrounding charges. If A is not initially connected to A, its potential is equal to the superposition of the electric potential generated by the charges on the surface of A.

Once A is internally connected to A, A's potential rises (the metal potential is equal everywhere) and can only be helped by A's charge. If there is an extra strong positive charge body next to A, the positive charge on A will be forced back, and even negative electricity will be generated. Thanks, got it?

Option B: Yes.

Analysis: When the B ball is in contact with the inner surface of A, A, B, C and the wire become a conductor, obviously the net charge of the positively charged conductor is only distributed on the outer surface when the electrostatic equilibrium is present, and this outer surface is the outer surface of A and the surface of the C ball, and they become equipotential bodies. Therefore, when the B ball leaves the inner surface of A and mentions the position in the diagram, there is no longer a flow of charge, i.e., the B ball is uncharged and the C ball is positively charged.

Note: Since the C sphere is positively charged, the outer surface of A has less positive charge.

Don't think about it too complicated.

The outside of the sphere is positive, and the inside of the sphere is negatively charged, and the same sex is repulsed, so B has a negative charge.

No matter how far away, B is connected to C, so C is positively charged.

Question 1: The opening angle of A The opening angle of B does not change, because C touches the inside of A, there is no transfer of charge, the electric potential on A does not change, and the opening angle does not change.

Second question: The opening angle of A becomes smaller, and the opening angle of B becomes larger. Until finally A becomes 0 and B becomes maximum, because C touches the outside of A, a transfer of charge occurs, the electric potential on A becomes smaller, and the opening angle decreases.

b has an electric charge and b has an opening angle. Finally the charge on A is completely transferred to B.

Question 3: The opening angle of A becomes smaller, and the opening angle of B becomes larger. If A and B are exactly the same shape, the last angle of A and B is the same. The difference is that the second question is that a will not be 0 in the end.