High 2 Physical Charge Please answer in detail, thank you 1 20 26 3

This is very simple, with the way of induction of electricity, the specific method is as follows: let BC contact, and arrange it in a straight line with A (not in contact) near A, so that due to electrostatic induction, the free electrons in B and C are affected by positively charged A, negative electricity appears near A, and positive electricity appears in places far away from A, and then B and C are separated, then B and C have different charges.

Let BC touch, A is close to the B terminal (away from the C terminal), first separate BC, and then move away from A, B will be negatively charged, and C will be positively charged "equally".

If A is near the C terminal, C is negatively charged and B is positively charged.

There's ball A on the left, and ball B on the right, and it's not in contact, so the left side inside the ball is negatively charged.

The inside right side is the punctual point Then, take the C ball. Touch the b-ball to the left so that the negative charge on the left reaches the c-ball.

Remove the A ball ,,B ball because there are fewer electrons. Be punctual. Can.

Stick BC together, then bring A close to B, (so that the 3 balls are arranged in a straight line in the order of A, B, C but A is not in contact with B), and then move C out of the way, then C will be positively charged and B negatively charged.

Move A out of the way again.

If B is brought closer to A, the outer surface of the side away from A will be positively charged because the charge induces B, so that C is in contact with that side, the free electrons of C will move to B and become positively charged, and C will be negatively charged due to the excess electrons, and B C will be separated from each other.

Because it is insulated, it is possible to electrify it by induction.

To make B positively charged, only A and B need to contact (contact live).

Let C be negatively charged, you should erect C with an insulating bracket, put A near C without contact, then the end of C close to A will be negatively charged, and the end away from A will be positively charged, and then continue to keep A in place, use another finger to touch the end of C away from A to draw the positive current away (in fact, let the positive current from other places such as the earth into C), and then let go of the contact of the hand, and then take away A, then C will be negatively charged.

e1＝kq1/r^2

e2=-kq2/r^2

Another e1 = e2

Get kq1 r 2=kq2 r 2

Then r 2 r 2 = q2 q1

If the combined field strength = 0, it is on the outside of the one with a smaller charge.

If the combined field strength is not = 0

This is in the middle of the two c

c has one in the middle, and the combined field strength is 2e

There is one outside the direction of the smaller charge, and the combined field strength is 0

Here is the issue of electrostatic induction. Think broadly. The B conductor is first induced by electrostatic electricity with an equal amount of dissimilar charge at both ends, and after grounding, the B conductor becomes a whole with the earth.

That is, into a large conductor. The B conductor becomes near A, which induces a different charge (negative charge) from point A, and at the far A end (you can't see it), the other side of the earth induces the same charge (positive charge) as A. So let's talk about the back b is only negatively charged.

The answer is not true, both the left and right sides of conductor b will have a negative charge.

The answer is not true, probably the left and right sides of the conductor will have a negative charge.

No, because of the positive charge, conductor B is still electrified.

For the first time, the A-ball balances at x. It is balanced by gravity, electric field force, and rope tension.

The red line represents these three forces, and the triangle formed by the thick black line is the triangle.

Similar triangles, there are:

mg F electro = l x

mg F electro = l x

The A-ball balances at x 2 for the second time. It is balanced by gravity, electric field force, and rope tension.

The thick blue line represents these three forces, and the triangle formed by the thin blue line is.

Similar triangles, there are:

m'g f electric' = l x 2

x is half of the original, and the electric field force is 4 times that of the original.

m'g = 8 f electro * l x = 8 mg

Select: C<>

by w=qu. Since s is the point of zero electric potential, u is the electric potential. Therefore, it is easy to find the electric potential of m.

According to the meaning of the topic, the work from s to n can be found. So the electric potential of n can also be found.

You said it yourself:

In the second sentence: 1, ep can be changed 2, the magnitude of ep is relative to zero potential energy (point surface). That is, the premise of your change EP is that you change the zero potential energy (point surface), and your zero potential energy (point surface) has changed, and the electric potential has naturally changed.

In the first sentence: the electric potential is determined by the nature of this point in the electric field, and has nothing to do with the tentative charge itself. It is not said that it has nothing to do with the selection of zero potential energy (point surface).

In summary: The only premise is that the zero potential energy (point surface) is constant. This doesn't need to be explained, so these two sentences are not contradictory.

In addition, electric potential is defined as: the electric potential energy possessed by a unit charge at a certain point in the electric field. The electric potential is only magnitude and directionless, it is a scalar quantity,-- and its value does not have absolute significance, but only relative significance.

You've studied mechanics, remember the gravitational potential energy, we usually think of the sea level as the gravitational potential energy is zero, above the sea level is positive, below is negative. However, in specific research, some convenient surfaces will be used as zero potential surfaces, such as sea level will not be selected as zero potential surfaces in classroom experiments.

The energy possessed by an object due to being lifted is called gravitational potential energy, and the magnitude of gravitational potential energy is related to two factors: mass and position, and the amount of change in gravitational potential energy is related to the work done by gravity, and has nothing to do with the selection of the reference plane

By analogy to the electrostatic field, the electric potential is relative, and the point potential of the electric field strength is not necessarily zero, and the electric potential along the electric field line must be reduced

The formula for finding the electric potential is e q, this is its definition, e to find by yourself, high school generally only find the uniform strong electric field, field strength (e) along the electric field line direction of the distance (d) q = ep, this ep divided by q, is the potential difference between the two points, if you find the potential difference between a point and the point where the electric potential is zero, you can know the potential of this point (the essence is ep q).

You can think of it as being only about the field charge and its position, and the potential is described with the help of a zero potential surface, for example, if the zero potential surface is selected, then the potential is deterministic and invariant, and it has nothing to do with the tentative charge.

You can refer to gravitational potential energy.

The gravitational potential energy of an object is related to the position of the object.

However, when determining the gravitational potential energy magnificence, a reference plane is selected.

Potential energy is the energy that two interacting objects have due to their position. In a system composed of two objects. Potential energy has a relationship with position. When we want to determine the potential energy magnitude, we choose the reference position because the position is relative.

b, the total charge number of a and b is conserved, and the number of pronuclei per metal ball can be changed.

a. Electrified and unelectrified.

b Negatively charged.

- Charge energy shift, negative electrical energy shift.

The force between the contact charges of ab moves the negative charge of the b ball to a, the total amount of charge is conserved, and the number of pronuclei is conserved.

a. Electrified and unelectrified.

b Negatively charged.

- Charge energy shift, negative electrical energy shift.

The interaction between the contact charges of ab is the migration of the negative charge of the b ball to a, the total amount of the charge is conserved, the number of pronuclei does not change, and the pronucleus is transferred.

Unchanged and prokaryotic metastases. Select B

First of all, the force on the two particles is equal, so the acceleration of b should be a 2, and the acceleration of b is half of the acceleration of a at all times, and the time is the same, so the velocity of b is v 2, and the potential energy lost is the sum of the kinetic energy of ab: 3mv square 4

A: False. B: False.

C: False. D: Right.

I'm giving you the answer so late ... Add a few points