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1、2021/3/9,1,2.2 Electric Field , Electric Field Intensity,1. Electric Field,(a) （without medium）,charge charge,(b) （ether）,(c) electric field,Question How does the interaction force between charges transmit?,Three viewpoints,(1832,Faraday),2021/3/9,2,Electric field is a special matter distributes rou
2、nd charge.,A region of space characterized by the existence of a force generated by electric charge .,Definition,2021/3/9,3,the field distributed round a static charge is called the electrostatic field, and the static charge is just called the source charge of electrostatic field.,Coulombs force is
3、really electrostatic field force.,2021/3/9,4,+,What force will a positive “test” charge feel if placed into the electric field?,2021/3/9,5,2. Electric Field Intensity,(1). Test Point Charge,Purpose,The test particle is used to measure the force and thus detect the existence of the field and evaluate
4、 its strength.,2021/3/9,6,The charge as detector should be a point charge whose electric quantity is smaller far than the source charge in order to decrease the disturbance as much as possible to the source charge and its electrostatic field.,Requirement,7,2. Definition,Unit N/C, V/m,At a point in t
5、he given electrostatic field the ratio of the electric field force on the test point charge with its electric quantity,is not related with value of , and is called the electric field intensity at the point in electrostatic field.,2021/3/9,8,Discussion,Magnitude the electric field force on unit posit
6、ive charge,Direction,Unit N/C 、V/m,the uniform electric field, non-uniform electric field,the static or steady electric field, the non-static electric field,is present whether or not we introduce a test particle into the field.,2021/3/9,9,Vector.,2021/3/9,10,3. Superposition Principle of Electrostat
7、ic Field,The total electric field intensity at a point in space equals the vector sum of the electric field intensity.,2021/3/9,11,4. Calculation,(1). Due to the System of Point Charges (The charges are countable ).,2021/3/9,12,The electric field at point P due to a group of source charges can be ex
8、pressed as,where ri is the distance from the ith charge qi to the point P (the location at which the field is to be evaluate ).,is a unit vector directed from ri toward P.,2021/3/9,13,(2). Due to Continuous Charge Distributions,P,Element Analysis Method,is a vector directed from the source charge to
9、ward the field point.,2021/3/9,14,The integral is a vector operation and must reduce it to a scalar integral.,The integral is over all the charge creating the field.,2021/3/9,15,Charge Density,2021/3/9,16,Reduce the vector integral to a scalar integral,2021/3/9,17,Example 1 Electric Field of a Dipol
10、e,An electric dipole consists of two point charges Q and Q separated by a distant of l, as in figure.,Many neutral atoms and molecules behave as dipoles when placed in an external electric field.,Many molecules, such as HCL, are permanent dipoles. (H+ ion , CL-ion.),The product is called electric di
11、pole moment.,2021/3/9,18,The electric field at the point P, perpendicular bisector to the dipole axis.,Question,2021/3/9,19,r l,r+= r- r,2021/3/9,20,Example 2. The Electric Field Due to a Charged Rod,P,X,a,l,A rod of length l has a uniform linear charge density and a total charge Q. Calculate the el
12、ectric field at a point P along the axis of the rod, a distance a from one end.,2021/3/9,21,choose an element of the charge distribution.,Express the charge dq of the element in terms of the other variables within the integral( in this example, there is one variable, x). Use dx to represent the leng
13、th of one small segment of the rod. The charge on the small segment is,The field intensity due to this segment at the point P is in the negative x direction, and its magnitude is,2021/3/9,22,Each element of the charge distribution produces a field in the negative x direction, and so the vector sum o
14、f their contributions reduces to an algebraic sum.,Discussion,If the point P is far from the rod( ),then l in the denominator can be ignored, and,This is just the form of a point charge.,2021/3/9,23,Example 3,A ring of radius R has uniform positive charge per unit length, with a total charge Q . Cal
15、culate the electric field at a point P on the axis of the ring at a distant x from the center of the ring.,The Electric Field Due to a Uniform Ring of Charge,（2）R x,（1）,Discussion:,2021/3/9,25,Example 4,Below figure shows a circular plastic disk of radius R that has a positive surface charge of unif
16、orm density on its upper surface. What is the electric field at point P, a distance x from the disk along its central axis?,The Electric Field of a Uniformly Charged Disk,2021/3/9,26,5. Sum-up: Calculating the Electric Field,Problem-solving strategy:,Analysis the distribution of charges.,Applying th
17、e superposition principle of the field intensity.,If the charges are countable, the resultant field is the vector sum of the fields due to the individual charges.,when confronted with problems that involve a continuous distribution of charge, you must divide the charge distribution into infinitesima
18、l pieces, and the vector sum is carried out by integrating over the entire charge distribution. (Element Analysis Method),2021/3/9,27,Symmetry,Whenever dealing with either a distribution of point charges or a continuous charge distribution, take advantage of any symmetry in the system to simplify your calculations.,