An uncharged, hollow metal ball with a little charge q, 5 inside the ball

First of all, there are two types of electrostatic shielding. Outer shielding and full shielding.

If there is a charged body in the cavity conductor, an equal amount of induced charge will be generated on its inner surface when the electrostatic equilibrium. If the shell is not grounded, the outer surface will produce an induced charge of the same amount and the same number as the internal charged body, and the electric field of the induced charge will have an impact on the outside world. If the shell is grounded, even if there is a charged body inside, the algebraic sum of the charge induced on the inner surface and the charge on the charged body is zero, and the induced charge generated on the outer surface flows into the ground through the ground wire.

The outside world cannot affect the inside of the shell, and the influence of the internal charged body on the outside world is also eliminated, so this kind of shielding is called full shielding.

Regardless of whether the closed conductor shell is grounded or not, the internal electric field is not affected by the charge and electric field outside the shell. The electric field outside the shell of a grounded closed conductor is not affected by the charge inside the shell. This phenomenon is called electrostatic shielding. d

Answer: A Analysis: The movement of the charge inside the ball will affect the distribution of the electric field inside the ball. The induced charge inside is equal to the charge in the spherical shell, and it can be seen from Gauss's theorem that it does not affect the induced charge distribution on the outer wall of the spherical shell, and therefore does not affect the electric field at each point outside the sphere.

b, when on the outside, the charge is shielded by the star.

But the answer upstairs is questionable.

It is not because the electrons produce a changed electric field, but because of the change in the distribution position of the point charge that causes a change in the electric field inside the sphere.

Is it a c-pull.

It should be electrostatically balanced, and the internal and external field strength should not change, which is an equipotential body.

Why don't you take it to your physics teacher, he will definitely praise you for being so studious!

The field strength inside the metal ball is 0, that is, inside the metal ball, the field strength generated by the induced charge should be in the opposite direction to the field strength generated by q.

So you can tell that ACD is correct and B is wrong.

A. According to the formula of the point charge field strength, the field strength of the point charge Q at the center of the sphere can be obtained e=kqr, and the direction is horizontal to the left, so A is correct

B, because the metal ball is in electrostatic balance, the closing field strength of point 0 is zero, so B error C, after the electrostatic equilibrium, the closing field strength in the metal ball is zero everywhere, the additional electric field generated by the induced charge on the metal ball and the field strength generated by the point charge q are small, the direction is opposite, cancel each other The small direction of the field strength generated by the point charge Q at each point in the ball is different, then the field strength generated by the induced charge on the metal ball is different in various places in the ball, so C is correct

d. Because the combined field strength at the center of the sphere is 0, the field strength of the induced charge on the spherical surface at the center of the sphere o is small e = e = kq

r, the direction is horizontal to the right, so d is correct

Therefore, ACD

A point charge +q is placed in the spherical shell, and the spherical shell is electrostatically induced under the action of the electric field of q, and when the electrostatic equilibrium is reached, the inner surface of the spherical shell induces an equal amount of -q, and the net charge of the spherical shell is zero, so the outer surface of the spherical shell must be induced charge +q - the following explains why the induced charge on the outer surface is always evenly distributed and does not change with the change of q position in the spherical shell:

First of all, these induced charges are simultaneously affected by the induced charge on the inner surface of the spherical shell and the action of q in the spherical shell, because the induced charge -q on the inner surface of the spherical shell and the +q algebraic sum in the spherical shell are the charges and zero in the spherical shell, therefore, the induced charge on the outer surface of the spherical shell is offset by the induced charge -q on the inner surface of the spherical shell and the charge +q in the spherical shell, so the induced charges on the outer surface of the spherical shell only interact with each other, and this mutually exclusive effect makes them stay away from each other and occupy the maximum area of the outer surface of the spherical shell as much as possible. The end result of this action can only be,Evenly distributed on the outer surface(As long as it is not uniform, where the charge density is large, the repulsive force of the charge is greater than that of the charge elsewhere, and it continues to move, and finally the equilibrium is evenly distributed).

a. The metal sphere is placed in the electric field, and the electrostatic induction phenomenon occurs, and finally it is in a state of electrostatic equilibrium, and the field strength inside the metal ball is zero everywhere, so A is correct;

b. The density of the electric field line indicates the magnitude of the field strength, and the electric field strength of point A is greater than that of point B from the image, so B is correct;

The electric field line where point C and A are located starts from Q and terminates with the uncharged metal ball, so the electric potential of point A is higher than the electric potential of the metal ball, and the electric field line at point B is emitted from the metal ball to infinity, so the electric potential of the metal ball is higher than the electric potential of point B, that is, the electric potential of point A is higher than that of point B, so C is correct;

d. The higher the potential, the smaller the potential energy of the negative charge, that is, the potential energy of the negative charge at point A is smaller than that at point B, so D is wrong

Therefore, ABC

A: The ball is not charged; Grip and dismantle.

Because a positively charged ball is placed into the original uncharged metal cavity spherical shell, the same charge repels each other, and the different charges attract each other, then the negatively charged electrons on the metal shell are attracted to the metal ball by the positively charged ball, and the positive charge on the metal ball.

Neutralization, the metal anti-withering conductor is not charged inside, and the outer surface is left with a positive charge, so the outer surface is positively charged; That is, the charge distribution of the original uncharged metal spherical shell has changed, and the electrons move to the inner layer, neutralizing the charge of the ball, and the outer surface of the metal is positively charged because of the lack of electrons, so the conductor in the electrostatic equilibrium state, there is no charge inside, and the charge is only distributed on the outer surface, then the ball is not charged, and the outer surface is positively charged

Put the positive charge inside the metal spherical shell, due to the action of electrostatic induction, a negative charge will be induced on the inner wall of the spherical shell, and a positive charge will be induced on the outer wall of the spherical shell, so between the shell walls of the metal spherical shell, due to the joint action of the electric field of the positive charge and the induced electric field, the electric field strength between the shell walls of the metal spherical shell is zero, so there is no electric field line between the shell walls of the metal spherical shell, and the electric field distribution outside the spherical shell will not be affected, so the correct one is b

Therefore, choose B

Since the electric field line at A is dense, the electric field line at C is sparse, and the field strength at B is zero, then EA EC EB

According to the decrease of the electric potential along the direction of the electric field line, the whole metal spherical shell b is an equipotential body, and the surface is an equipotential surface, and the analysis shows that the electric potential relationship is ua ub uc

So the answer is: EA EC EB; ua＞ub＞uc

Put a ball with positive charge q into the original uncharged metal cavity ball shell, the negatively charged electrons are attracted to the inner surface by the positively charged ball, the inner surface is negatively charged, the outer surface is left with a positive charge, the outer surface is positively charged, and R is in the electric field, the phenomenon of electrostatic induction occurs, which leads to the appearance of an induced charge on R, if the ball with a positive charge Q is put into the cavity with -q charge m, because the same charge repels each other, and the different charges attract each other, then there is no excess charge on the metal shell, then there is no electric field at P, Therefore there is no induced charge, so abc is wrong, d is correct

Therefore, d