Several physical questions about electric fields, physical problems related to electric fields

Personally, I understand it.

1. The field strength at the midpoint is the smallest on the isometric heterogeneous charge line.

This sentence, please look at it in combination with the picture.,It's very easy to understand.,I can't send **.,Give you a connection.,There's a picture in it.。 Or you can try to draw the electric field lines yourself, the electric field lines at the midpoint are sparser than those at both ends, and the electric field strength is directly proportional to the density of the electric field lines. Keywords: density of electric field lines.

2. The field strength at the midpoint is the largest on the mid-vertical line.

Again, if you look at the diagram, on the mid-perpendicular line, are the electric field lines more sparse at the point that is farther away from the midpoint you are talking about? Then the field strength of the point farther away from the midpoint must be small, and the field strength of the midpoint must be the largest because the vertical line is symmetrical from side to side. Keywords: density of electric field lines.

If I have something that I can't say clearly, I don't have to dwell on it, just understand the stem. The topic of electric field is more abstract, it would be nice to see more and do more - hopefully

Here you have to look at the scope of the comparison.

The field strength at the midpoint is the smallest on the isometric xenogeneous charge line".

Of all the points on the xenocharge line, the lowest field strength is the midpoint of the line.

The field strength at the midpoint of the midvertical line is the largest".

Of all the points on the perpendicular line, the largest field strength is the midpoint of the perpendicular line, which is the midpoint of the heterogeneous charge line in the above sentence.

To sum up, the field strength of the intersection point of the perpendicular line and the charge line is the smallest when compared to the field strength of the other points on the line.

It is the largest when compared to the field strength of other points in the perpendicular.

That is, this field strength relationship: the point on the line except for the intersection "the intersection point" except for the point on the vertical line of the intersection point.

First of all, from 4qe = 10mg, the field strength E can be obtained. Then, from the kinetic energy theorem, the work done by gravity is 10mg*3l-the work done by the electric field force qe*(3l+2l+l)=1 2*(10m)v*v can be calculated.

First of all, from the change in the direction of the electric field, we know that the distance of the small object moving for the first time is farther, which means that the friction on the horizontal plane is less for the first time. So we know that the ball carries a negative charge.

According to the law of conservation of energy, the negative work done by friction is equal to the kinetic energy of a small block. Let the coefficient of friction be , so:

For the first time, the electric field force is upward, and the magnitude of the frictional force is: (mg-eq) according to the conservation of energy: 1 2mv 2=(mg-eq) ab

The second time the electric field force is downward, and the magnitude of the frictional force is: (mg+eq), according to the conservation of energy: 1 2mv 2=(mg+eq) ac

It is possible to solve from the above two equations. e=mg/3q

First of all, the second question, which can be considered from the conservation of energy, the positive charge from the A side to the C plane, the kinetic energy is zero, then all the electric potential energy is converted to electric potential energy, then the difference between the potential energy of the positive charge on the C plane and the potential energy on the A plane is 20j

In the present problem, let b be the zero potential energy surface, and it can also be seen that the kinetic energy on the b side is 10j, and the total energy is 10j, then we can know that the answer is 8j

Let's talk about the fourth hoof.

At the beginning, two charges of the same kind are infinite, which means that there is no interaction force between the two, that is, the electric potential energy is zero, and later, b moves towards a, because the two are the same kind of charge, and with the addition of distance, there is a force that repels each other, and as the distance is added, electric potential energy is generated between the two, so the value obtained by adding the kinetic energy of a and b must decrease with the increase of electric potential energy, and the minimum value of the addition of the kinetic energy of the two is the minimum value of the addition of the kinetic energy of the two, that is, when the distance between a and b can no longer be reduced, that is, the two velocities are equal (think about it yourself, you will understand, otherwise, just look at the formula.) You'll never learn physics well) so keep 4mv=mv2+4mv2 according to momentum

v2 is what you are asking for.

Knowing the velocity means knowing the kinetic energy, according to the conservation of energy, the next question can be solved.

Finally, there is the fifth question.

It should be said in the question that the point q does not detach from the inclined plane, and the fixed positive charge at the midpoint must have an upward force on the point of particle, so this question must be calculated by gravity (there is no other force that makes it downward), if only the electric potential energy is considered at d and c, the electric potential energy is actually equal (at the same equation surface), so from d to c, it is only the gravitational potential energy that is converted into kinetic energy.

So... Ball it yourself, already said very clearly.

It's a matter of field strength. It is easy to obtain eq=, mg=. The resultant force is f=, and the angle between the direction and the vertical direction is 37 degrees.

You can rotate the diagram 37 degrees clockwise, and see what, the resultant force f is vertically downward, that is, it represents the sa, and now the force eq and mg of the two fields are combined into a force f, and it is the familiar vertical direction of the circle luck to take the Kai defeat. So just consider the resultant force, and think of f as the gravitational force g of the new gravitational field'(You can also find the new gravitational acceleration g.)', but it is not necessary).Then the ball moves from the bottom to the vertical top (that is, the direction of the rope of the original figure that does not rotate and the focus of the circle), and the gravitational potential energy increases f*2r=; Since the rope is soft (not rigid like a lever), the ball must have a minimum velocity around the highest point, satisfying m*v 2 r=mg'=f, then the kinetic energy required is e kinetic = (1 2)*m*v 2=(1 2)*r*f=, then the total energy required is e=.

I don't know if you understand, it will be the same to add magnetic fields in the future, it is all a matter of field strength.

Because it is a point charge, we can use Coulomb's law: F=KQ1Q2 R2 Let the charge of a be 4q, and the charge of b is q, which is 5cm apart (, and the repulsion force is so 4kq 2

kq 2 = 20 cm apart (4 kq 2

Bring in kq 2= and get f=

Since kq 2=, and k is a constant = 9*10 9n ·m 2 c 2, it is treated as 9*10 9

I bring it in and get it.

q1=(1/3)*10^-6=b

q2=(4/3)*10^-6=a

r increased to 4 times f 1 r 2

f(20)=

f=kq*4q r 2=4kq 2 r 2b amount of electricity q = (fr 2 4k) = (1 2) (fr 2 k)a amount of electricity 4q = 2 (fr 2 k).

According to the conservation of energy and the distribution characteristics of the electric field (if you don't know if you don't know the socks and pants, you can go here to see the dense electric field lines at point A, which means that the electric field strength is large, the potential energy is large, and the point B is relatively small. The electric potential energy at point A is large, and the kinetic energy at point B is large.

A: Can't judge the positive or negative charge (look at the ** given to you above and you will understand) BThe direction of acceleration is the direction of the force, which must be along the electric field line and pointing to the left side of the graph.

c.It is not possible to judge the positive or negative of work (the direction of motion is unknown).

d.If the energy is conserved, the potential energy at point b must be converted into kinetic energy in order to balance the pure disadvantages. Therefore, the kinetic energy of point B is large.

ABC is like the forest banquet upstairs, if the electric field from A to B dismantles the force to do negative work, the electric potential energy increases, because the energy is conserved, so the kinetic energy of B is small and the potential energy is large, and the electric field force from B to A does positive work, and the potential energy of B is large and the kinetic energy is small, so it has nothing to do with the direction of motion.

Taking the first quadrant as an example, the electric field line of B2O is perpendicular to B2O to the left, and the electric field line of ob1 is perpendicular to ob1 upward, and when superimposed, it becomes oblique 45 degrees to the upper left, and each electric field line is parallel to each other, and the equipotential surface is perpendicular to the electric field line, and then it becomes like this.

Analysis: The distribution of the equipotential surface in the area can be judged according to the field strength of a certain quadrant, such as the horizontal charged plate in the fourth quadrant is negatively charged, the electric field is vertically upward, the vertical plate is positively charged, the field strength is horizontal to the right, and the field strength is a vector, so the direction of the combined field strength is obliquely to the upper right and then the equipotential surface is determined to be oblique to the lower right according to the electric field line and the equipotential surface Answer: Solution:

The electric field line is terminated from the positive charge to the negative charge or infinity, or from infinity to the negative charge, in the fourth quadrant, the electric field generated by the positive charge is horizontally to the right, and the electric field generated by the negative charge is vertically upward, so the combined field strength is obliquely to the upper right, that is, the electric field line is obliquely to the upper right and the electric field line and the equipotential surface are perpendicular to each other, so the equipotential surface should be obliquely to the lower right Therefore, only A is correct, so A is selected

Finding a zero electric potential surface (a point with zero electric potential at the same distance from two planes) tells that the answer is a.