Basic essentials 1. Electric field Faraday believed that there is a force generated by it around the charge, and the force exerted on other charges in the electric field is given by this electric field. Modern physics believes that electric field and magnetic field are an objecti

Basic essentials

1. Electric field

Faraday believes that there is a force generated by it around the charge, and the force exerted on other charges in the electric field is given by this electric field. Modern physics believes that electric field and magnetic field are an objective existence and are related to each other. They are collectively called . Field and magnetic field are two different forms of material existence.

2. Electric field intensity

The ratio of the test charge at a certain position in the electric field to its value is called the electric field intensity, expressed by the formula E=; the unit is N per library, and the symbol is N/C; the electric field intensity is a vector, and the direction is The direction of force is the same.

3. The electric field of point charge The superposition of electric field intensity

. The field strength formula of the electric field formed by a point charge in a vacuum is: E=k, where k is the electrostatic force constant and Q is the charge amount of the field source charge.

. If the field source is multiple point charges, the electric field intensity at a certain point in the electric field is the vector sum of the electric field intensities generated by each point charge alone at that point.

test point analysis

question type 1: electric field intensity calculation and significance

identification mark: electric field intensity , electric field intensity meaning

mental formula: electric field intensity is a vector, and the electric field source distance is fixed. The force field exerted by small q has nothing to do with the force exerted by the unit charge. Three major formulas should be memorized to calculate and distinguish point charges.

analysis and application:

The electric field strength is a vector, and its size and direction are determined by the size and distance of the (generated) field source charge, and have nothing to do with the (test) charge put into it.

What the electric field intensity represents (meaning) is the electric field force experienced by the unit charge, which represents the mechanical properties of the electric field. There are three main relationships for calculating electric field strength! We must first distinguish whether it is the uniform electric field or the point charge electric field! ! !

question-taking routine: electric field strength calculation and significance !

. Introduce different test charges at and in the electric field, and obtain the relationship between the electric field force on the test charge and the change of the charge amount as shown in the figure, then

A. B． C. D. Cannot determine the size relationship between ,

Exercise:

. In order to measure the electric field intensity excited by the charge at a point, put in a test charge and measure the force, then

A. The test charge can only be positively charged B. If the electric charge is large enough to affect the distribution of the electric charge, it cannot be used as a test charge C. If a test charge is replaced at the point and the measured force is, the conclusion will be established D. Move the test charge farther away, and the conclusion will be established.

, There are several different understandings of the field strength formula of point charges below. The incorrect one is

A. In the electric field of a point charge, the field strength at a certain point is directly proportional to and inversely proportional to B. At that time,; at that time, C. In the electric field generated by a point charge, the direction of the field strength at each point must be away from the point charge D. With the point charge as the center and the radius as the sphere, the field strength is the same everywhere

. As shown in Figure A, there is a point charge on the axis (not shown in the figure), , and are three points on the axis. , the relationship between the electric field force on the test charges at two points and the amount of charge carried by the test charges is as shown in Figure B, then

A. The electric field intensity at the point is B. The electric field intensity at the point is C. The point charge is between D. The point charge is between

. The negatively charged test charge, in a certain electric field, is only affected by the electric field force and can complete the following two kinds of motion: ①Move on the electric field line, ②Do uniform circular motion on the equipotential surface. The electric field may be caused by

A. A positively charged point charge forms B. A negatively charged point charge forms C. The formation of two separate point charges with equal amounts of positive charge D. Two separate point charges with equal positive and negative charges form

. As shown in the figure, a metal ball with an electric charge of is fixed on an insulating bracket. At this time, the electric field intensity at the outer point of the ball is. When a charge of When a point charge of is placed on a point, the measured electrostatic force on the point charge is; when a point charge of electric quantity is placed on a point, the measured electrostatic force acting on this point charge is , then in international system of units

A. The value is equal to B. The value is equal to C. The smaller it is, the closer the value of is to D. The smaller the ratio is, the closer the value of is to

. As shown in the figure, there is a point in the vacuum with a point of charge, and there are two points in the electric field it generates. The field strength of the point is, and the direction is at an angle with the connecting line. The field strength is , and the direction is at an angle to the connecting line.Regarding the relationship between the field strength of two points , and , , the following conclusion is correct:

A. B． C. D.

. From the definition of electric field intensity, it can be seen that at the same point in the electric field

A. The electric field strength is directly proportional to and inversely proportional to B. No matter how the amount of electricity carried by the test charge changes, it remains unchanged C. The electric field force experienced by the charge at a certain point in the electric field is large, and the electric field intensity at that point is strong D. The electric field force experienced by an uncharged small ball at point P is zero, then the field strength at the point must be zero

. As shown in the figure, in a uniform electric field, a charged small ball with mass and charge is given by When released at rest, the trajectory of the charged ball is a straight line, and the angle between the straight line and the vertical direction is. The gravity of the ball cannot be ignored, so the field strength of the uniform electric field is

A. The only value is B. The maximum value is C. The minimum value is D. The minimum value is

field strength	(field source) point charge?	Known conditions	Charge Electricity	Electric field lines?	function changes
A
B	Examples: , field source charge , is a positive point charge. Check that the charge , is a negative point charge. They are separated by , and both are in vacuum, as shown in the figure. Asking for: the electrostatic force received by (1). The field strength at the point where (2) is located. (3) If is replaced by the positive point charge of , then find the force on and the field strength at the point. (4) After removing the test charge, find the field strength at the point.	. Put a point charge on a point in the vacuum, a straight line passes through the point, a distance of , put a point charge on the point, as shown in the figure. Find: the force exerted by (1) at point. The field strength at (2) point. (3) Take away the field strength of the point. (4) After taking it away, which field intensity will be greater?
0. The distribution of electric field lines and in a certain area is as shown in the figure. The middle electric field line is a straight line. A positively charged particle starts from a point on the straight line from rest and moves to a point under the action of the electric field force. Take the point as the origin of the coordinates, the positive direction of the axis along the straight line to the right, and the gravity of the particle is ignored. Then during the movement, the following graphs about the change of particle movement speed and acceleration with time, the particle kinetic energy and the change of electric potential with displacement on the movement track may be correct: A. 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