重邮大学物理英文版PPT4

1、2021-10-2512.2 electric field , electric field intensity 1. electric field (a) （without medium）charge charge(b) （ether）charge ether charge (c) electric fieldquestionhow does the interaction force between charges transmit? three viewpointscharge field charge(1832,faraday)2021-10-252electric field is

2、a special matter distributes round charge. a region of space characterized by the existence of a force generated by electric charge .definition2021-10-253 the field distributed round a static charge is called the electrostatic field, and the static charge is just called the source charge of electros

3、tatic field. coulombs force is really electrostatic field force. 2021-10-254+what force will a positive “test” charge feel if placed into the electric field?2021-10-2552. electric field intensity(1). test point charge purposethe test particle is used to measure the force and thus detect the existenc

4、e of the field and evaluate its strength.2021-10-256f0qthe 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 72. definition0q

5、fe unit n/c, v/mat a point in the given electrostatic field the ratio of the electric field force on the test point charge with its electric quantity f0qis not related with value of , and is called the electric field intensity at the point in electrostatic field. 0q2021-10-2580qfe discussion magnitu

6、de the electric field force on unit positive charge directionunit n/c 、v/mthe 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.e2021-10-259.cos ,cos , cos,222eee

7、eeeeeeezyxzyx ,d ,d ,dzzyyxxeeeeee )(),(rezyxee vector.2021-10-25103. superposition principle of electrostatic field nfffff321 002010 qfqfqfqfn niineeeee121the total electric field intensity at a point in space equals the vector sum of the electric field intensity.2021-10-25114. calculation (1). due

8、 to the system of point charges (the charges are countable ).eqr1 1r1q2f3f2q3q1 1fq03 32 21 1ffff0 0qfe0 03 30 02 20 01 1qfqfqf3 32 21 1eee2021-10-2512 iiee iiiirrq2004the electric field at point p due to a group of source charges can be expressed as where ri is the distance from the ith charge qi t

9、o the point p (the location at which the field is to be evaluate ). is a unit vector directed from ri toward p.oir2021-10-2513304dqderrqdqr(2). due to continuous charge distributionsep element analysis method is a vector directed from the source charge toward the field point.r2021-10-25143014qqreded

10、qrthe integral is a vector operation and must reduce it to a scalar integral.the integral is over all the charge creating the field.2021-10-2515201dlinear harge density 4lcdqdlerr，charge density201 surface harge density 4dscdqds err，201 volume harge densidity 4dvcdqdv err，2021-10-2516dqrxex3041reduc

11、e the vector integral to a scalar integralkejeieezyxdldsdvdq dqrredeqq20041 2021-10-2517example 1 electric field of a dipolean electric dipole consists of two point charges q and q separated by a distant of l, as in figure.l+-qqmany neutral atoms and molecules behave as dipoles when placed in an ext

12、ernal electric field.many molecules, such as hcl, are permanent dipoles. (h+ ion , cl-ion.)the product is called electric dipole moment.lqp lqpe2021-10-2518the electric field at the point p, perpendicular bisector to the dipole axis.lrreee+-qqquestion ly 2021-10-2519lrreee3 30 04 4rrqe3 30 04 4rrqer

13、eee3 30 04 4rrrq rrllrr3 30 04 4rlq 3 30 04 4rpee lqper lr+= r- r+-qq2021-10-2520example 2. the electric field due to a charged rod pxalea rod of length l has a uniform linear charge density and a total charge q. calculate the electric field at a point p along the axis of the rod, a distance a from

14、one end.2021-10-2521choose 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 length of one small segment of the rod. the charge on the small segment isdxd

15、q the field intensity due to this segment at the point p is in the negative x direction, and its magnitude is ed2xdxked 2021-10-2522each element of the charge distribution produces a field in the negative x direction, and so the vector sum of their contributions reduces to an algebraic sum.)(2laakqx

16、dxkdeelaa discussion if the point p is far from the rod( ),then l in the denominator can be ignored, and 2akqe la this is just the form of a point charge.2021-10-2523example 3 a ring of radius r has uniform positive charge per unit length, with a total charge q . calculate the electric field at a po

17、int 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 chargeldee0 0ll/decosdee 2 20 04 4rdqde eee/rx2 20 04 4rdq 3 30 04 4rdqxl xq2 23 32 22 20 04 4rxixqee/ ededed/（2）r x204xiqe0 x0e（1）discussion:dldedeeexcosdd 2021-10-2525example 4 belo

18、w figure shows a circular plastic disk of radius r that has a positive surface charge of uniform 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 disk2021-10-25265. sum-up: calculating the electric field problem-solving strategy:analysis the distribution of charges.applying the s