Fill in the blanks with electrostatic field, and the electrostatic field is .

I'm sorry, I only saw the problem tonight!!

Here's what I thought:

The 1, 2, and 3 balls connected by ropes are regarded as a system, and the system is not affected by external forces, so in the process of their respective motion, their centroid positions remain unchanged, that is, the center of the initial equilateral triangle, and each ball is fixed around the center for periodic motion.

The rope between 1 and 2 is cut and each ball starts to move, at first 1 ball moves to the upper left, 2 balls move to the lower left and symmetrically with the 1 ball about the horizontal line in the figure, and the 3 balls move in a straight line to the right in the horizontal direction. Note that each ball moves under the action of the electric field force and the constraint of the rope, and the change of their position must correspond to the change of the potential energy of the system, so it is not difficult for us to judge that the ball must move to the dotted line position in the diagram and return to the original position, and do periodic motion in these two positions.

Therefore, the maximum displacement of the ball 3 relative to the initial position should be <>

During the movement of the ball from one position to another, the potential energy of the ball is converted into its kinetic energy. So when is the ball getting the most velocity? Of course, it is when the potential energy of the system is minimal, when the three small balls move to the same straight line.

The maximum velocity of the ball comes out.

Another way of thinking can also come up with the maximum speed of the ball 3! From the very beginning of the movement to the first time when the three balls arrive on the same straight line, the combined force of the two ropes and the two teams of small balls 3 is to the right, so that it accelerates; After that, it's to the left, slowing it down. Therefore, when the three balls are on the same straight line, the velocity of the ball 3 is the largest.

Am I right? What is the answer to the second question?

Answer]: A The two basic equations of the electrostatic slag field are.

From this, it can be judged that Yeliang Nasong has no electrostatic field, which is an active and no rotation field.

1 There are no points with equal field strength in the electric field formed by the point charge, but there are two points with equal potential... That's right. The field strength is a vector quantity, which is distributed in a "sunlight-like" space, so there is no exact same field strength point. The electric potential is scalar, and the points on the "sphere" are all equipotential.

2 The positive charge must move from a high potential to a low potential only under the action of the electric field force... Wrong. When the initial velocity is not 0, it can move in the opposite direction.

3. A positive charge with zero initial velocity must move along the electric field line under the action of the electric field force... Wrong. It should be said "only under the action of the electric field force".

The differential relationship between the field strength e and the electric potential is 0 and is e=-gradv

The gradient of the electric potential) The potential is invariant and the gradient is 0

Landlord. Some symbols are hard to hit. I'm at a loss to take the text.

This question mainly examines the formula "electric potential energy = electric potential * electric charge".

a。Electric potential energy = electric potential * electric charge.

There are positive and negative charges. So with the same amount of charge, the lower the potential energy of the negative charge is at the higher the potential energy. The positive charge is reversed.

b。It's pretty much the same as item A. It's just that the a-variable is the electric potential. The B variable is the electric charge.

The key points are all in the positive or negative charge.

The landlord still looks at this formula: electric potential energy = electric potential * electric charge.

c。With zero electric potential energy at infinity as the pin shed, the electric field line with positive and negative charges obtains: the electric potential of the positive charge electric field is positive. The potential of a negative charge is negative.

So. The positive is always greater than the negative*

d.Similar to item c. Here is a negative sum and a negative number is positive. The positive is always greater than the negative*So the potential energy possessed by the positive charge must be less than that possessed by the negative charge.

When the initial kinetic energy is e, the time through the capacitor is t, the vertical moving distance of the capacitor is h, the vertical acceleration is a, the increased kinetic energy is e=uqh, h=1 2at 2, if the velocity is doubled, that is, t1=t 2, h1=1 2at1 2=h 4, the increased kinetic energy is e 4, and the initial kinetic energy is 4e, so the kinetic energy when flying out of the capacitor is 17 4e

Analysis: A: The electric field force and gravity of the ball A and the support force of the thin rod are not constant during the movement, so it is wrong for the ball to do uniform acceleration linear motion.

b: As soon as the ball falls, the electric field force does positive work on the ball, and the electric potential energy is converted into mechanical energy, and the mechanical energy increases. Subsequently, the electric potential energy does negative work on the ball, and the mechanical energy is converted into electric potential energy, and the mechanical energy decreases. b correct.

C: The electric potential energy of the ball at point B and point C is equal, so the mechanical energy of the two points is opposite, and the gravitational potential energy of point B is greater than point C, and the kinetic energy of point C is greater than point B due to the conservation of energy. c correct.

D: The electric potential of the ball is equal from point B to point C, the potential difference is 0, and the electric field force does not do work on the ball. Therefore, the electric field force of the ball from point A to point B and point A to point C is equal. d error.

I've been doing the math for a long time, and I think there's something wrong with your question. EB has some questions, isn't the graph a little wrong, because the calculated r has a root number, and I know that there is an original problem like this. This kind of question is very easy to calculate, and the second question is mainly to determine the position of q (remind you that the test charge is a small q, and the point charge is q!).

Remember! Otherwise, many people can't understand your title)

Let me briefly describe that if q is positive, then neither the left nor the middle can be put, because the direction of force will be negative to the x-axis, which does not match the meaning of the topic, it is possible on the right, but there will be no solution through calculation. Therefore it is impossible.

In the same way you can also push out other impossible situations, and finally determine that only in the middle can and must be negative. Don't be lazy, push yourself.

The answer is attached below.

1。According to the image, divide Q by F, ea = 40 eb = 2

2。The original question should be Here I will press you to paste the answer, you can understand the meaning, don't get entangled in the numbers, understand? Socks filial piety.

The ox axis coincides with one of the electric field lines of q, indicating that q is on the ox axis.

The electric field force of the tentative charge at points a and b is in the same direction as the positive direction of the x-axis. The charge at point A is positively charged and point B is negatively charged. Based on these conditions, it can be judged that:

q is negatively charged, somewhere between a b. Serving the gods.

Electricity of the same sex repels each other, and electricity of the opposite sex attracts. Assuming that q is positively charged and is on the left side of a, then the direction of force a is to the right and the direction of force to b is to the left, which contradicts what is known. Using a similar reasoning, it is not difficult to conclude that Q is between A B and is negatively charged. ）

a b 3 meters apart. Let q be r meters away from a and 3-r meters away from b. Rule.

ea = 40 n/c = kq/r^2

eb = n/c = kq/(3-r)^2

ea/eb = 16 = 3-r)^2/r^2

4 = 3-r)/r

4r = 3-r

r = m.

Substitution 40 n c = 9 10 9 n m 2 c 2 ) q ( m) 2

q = 10 (-9) coulombs.

Considering the electronegativity.

q = 10 (-9) coulombs.

Ans: q is negatively charged q = Coulomb. Between a b, a meter away. i.e. the coordinate position is meters.

ra＜r＜rb，e=qa/4πεr^2

Calculate the electric field strength r r1,e=0,Gaussian plane without charge r1 r r2,e=q1 4 r 2 r r2,e=(q1+q2) 4 r 2 electric potential u r r1,u=(q1+q2) 4 r,u=integral e*dl Here from infinity to the inner spherical shell r1 r r2,u=q1 4 r 2 here from infinity to between the inner spherical shell and the outer spherical shell r r2,u=0 here from infinity to the outer spherical shell, After the outer sphere is connected with a wire, the inner and outer sphere shell are equal potential, and the electric potential is 0

If you don't know, ask again; Satisfied! Good luck opening !!