I don t understand the concept of electric potential in the second year of high school, explain it i

To understand this way, it is important to remember that a positive charge in free motion will spontaneously run from a place with a high potential to a place with a low potential, while a negative charge will move in the opposite direction.

In the electric field generated by the positive charge, a positive test charge is put in, then the test charge will run farther and farther away from the low potential due to the repulsion of the same sex; Since the electric potential at infinity is zero, that is, the electric potential at the lowest point in the electric field generated by the positive charge is zero, the potential at each point in the electric field generated by the positive charge is positive.

For the negative point charge electric field, the positive test charge must move towards the negative point charge in the center no matter where it is placed, that is, the closer to the negative point charge, the lower the potential; If the distance is zero, then the potential of each point in the electric field generated by the negative point charge is a negative value less than zero.

In the electric field generated by a positive charge, the direction of the electric field line is pointed outward from the positive charge. Because the electric potential decreases along the electric field line, there is a minimum potential at infinity. If the electric potential at infinity is zero, then the electric potential elsewhere in the electric field is positive.

In the electric field produced by a negative charge, the direction of the electric field line is directed from the outward to the negative charge. Because the electric potential decreases along the electric field line, there is a minimum potential at the negative charge. The same leads to a second conclusion.

The electric potential gradually decreases along the direction of the electric field line.

The electric field generated by the punctual charge is outward, so the potential at the points around the punctual charge is positive; The electric field generated by the negative charge is inward, so the potential at each point around the negative charge is negative. (Specifying zero potential at infinity).

Taking the negative charge as an example, the electric field line it forms is directed towards itself, moving along the electric field line, the electric potential gradually decreases, and the electric potential at infinity is zero, then the electric potential near the negative charge is less than zero and is negative.

The case of a positive charge, on the contrary, is positive.

Analysis: Solving with the principle of potential superposition and integration.

Place the rod horizontally on the x-axis, and make the perpendicular line at the midpoint of the rod as the y-axis.

At point p, where the distance from the perpendicular foot (coordinate origin) is a, the electric potential is.

2 kr) dq and half the length of the rod for the integration interval (symmetrical).

dq is the amount of charge taken on a small length of the rod and r is the distance from dq to point p.

Get 2 k root number (x 2 a 2) ]dx

2 k 1 root number (x 2 a 2) ]dx

2 k ln[ x root number (x 2 a 2)].

Substituting the integral interval of x from 0 to l 2 into the above equation, the potential of point p is .

2 k ln{[ l root number (l 2 4* a 2)] (2 a)}

Note: When the rod is positively charged, it is positive, then the p-point potential is positive. When the rod is negatively charged, it is negative, then the p-point potential is negative.

In fact, these powerless concepts are much easier to understand by analogy with altitude in geography. First of all, we understand the concept of electric potential, which is similar to altitude, with zero electric potential and sea level, respectively, and the electric potential difference is similar to the problem of relative altitude. The generation of the electric field is also based on the existence of an electric potential difference.

In fact, the simple problem of electric field is very easy to solve, and some comprehensive topics involving electric field and magnetic field often make people scratch their heads and don't know where to start.

The requirements for high school are relatively low, and it is OK to understand the basic concepts and be able to do questions. After getting a question, first judge the scope of the question, and then judge the conditions given by the question, sort out what conditions the question gives, and what conditions are unknown, and what kind of formula and definite solution can be used to solve the answer to the question.

In short, practice makes perfect, read more solutions to classic problems, pay attention to the use of formulas and the direction of problem solving, and it will become more and more smooth to learn.

The important thing is not to look at the master, and it is best to rehearse a lot of things by yourself, so as to give you a deeper impression

The electric field is equivalent to the teacher's strength range "surrounding students", and the electric potential is equivalent to power, the teacher is in charge of the class cadres, and the class cadres are in charge of the students; The potential difference is that the voltage is equivalent to the poor ability, the ability of the teacher and the class cadres is poor, and the class cadres and classmates are poor.

If you understand the strength of the electric field, then you should understand the electric potential, one is the angle of the force and the other is the angle of the energy.

Compare them, and you'll get the idea, both in terms of definitions and physical meanings.

Take a look at this network diagram, I wonder if you will understand something?

In fact, the study of electricity is also based on mechanics.

There are also properties in electricity that describe the properties of force and energy.

e corresponds to acceleration a

Electric potential energy corresponds to mechanical energy in mechanics.

The electric potential decreases in the direction of the electric field lines (theorem).

Electric potential energy represents the work done on a point charge by the electric field force moving from that point to a place where the electric potential is zero. (concept). He is equal to the electric potential of the point multiplied by the value of the tentative charge.

The potential difference is the difference between two points of electric potential.

The denser the electric field lines, the greater the electric field strength.

The potential energy at this point is the difference between the potential at the point and the potential at zero multiplied by the amount of charge of this charge w=q*(u1-u2).

The electric field strength is equal to the force experienced by the charge at that point divided by the amount of charge e=f qFor a uniform electric field, the electric field strength of the two points is the difference between the electric potential at the two points divided by the distance between the two points e=(u1-u2) d.

If you have any questions, you can continue to ask.

According to the kinetic energy theorem w= ek

Here the electric field force and other forces do work, the kinetic energy increases, and the other forces do work, then the electric field force does the work, and the formula for the electric field force to do the work is w=qu, and u is the potential difference.

The solution is u=10 4v

The potential difference is 10 4V, because it is a negatively charged particle, and the electric field force does positive work, so the potential at point B is higher, and the potential of B is 10 4V

According to the kinetic energy theorem, it can be known.

The increase of kinetic energy is to do work with external forces, and it is obvious that the work done is other than the electric field force, and the title tells other forces to do work, from which it can be calculated that the work done by the electric field force is w=2*10 -5j

The formula for doing work using the electric field force is w=quab

UAB = 10000V is calculated

And UAB = a- b

Then b=-10000v

Conservation according to energy.

Find the electric field force to do work w=(8-6) (10 -5)jw=qu

The electric potential at point A is 0

Let the electric potential between A and B be UAB

uab=w/q

1. Let each segment be s, the two periods of time are S v1 and S v2 respectively, and then divide the sum of the two times by 2s, and S will be about 12m s

2. Let the two periods be t1 and t2, and the total displacement is 6t1 + 9t2, and divide by the total time t1 + t2 to get 8m s, then t1 t2 = 1 2 displacement ratio and so on 6t1 9t2 = 1 3

<> "I thought it was from high school, but when I found out, I became a sophomore."

Select A When a resistor is connected in parallel between AC, the resistance between AC will become smaller, the total resistance will be smaller, and the total current will become larger, so that the voltage shared by R2 and the internal resistance of the battery will become larger, so that the voltage between AC will become smaller, because point C is grounded, point C potential is 0, point B potential is equal to the voltage at both ends of R1, so point B potential increases, the voltage between point AC decreases, and point A potential is lower than point C potential, so point A potential is A=-UAC will increase.