電磁學(xué)第二版習(xí)題答案第一章（Chapterof the second edition of electromagnetism）

1、電磁學(xué)第二版習(xí)題答案第一章（Chapter 1 of the second edition of electromagnetism）Chapter I basic law of electrostatic field1.1 determine if the following statement is correct and explain the reasons.(1) a point of field intensity is the direction of the electric field force of the probe point charge.(2) the direct

2、ion of field intensity can be determined by E = F/q, where q can be positive and negative.(3) the field intensity generated by the point charge is the same everywhere on the sphere with the point charge.(1) x positive probe charge; (2) square root; (3) x is on the surface of the sphereE?The magnitud

3、e is equal.1.2 the surface of the hemisphere is evenly distributed with positive charge. How to use symmetry to determine the field strength of the ball?Answer: using the symmetry analysis, the components of the vertical axis cancel each other.1.3 is the following statement true? Why?(1) the total c

4、harge in the surface of the closed surface is zero.(2) when the total charge in the closed surface is zero, each point on the surface must be zero.(3) when the E flux of the closed surface is zero, the intensity of each point on the surface must be zero.(4) the E flux on the closed surface is provid

5、ed only by the surface charge.(5) the field intensity of each point on the closed surface is only provided by the charge on the surface.(6) under the conditions of gauss theorem, the distribution of charge is symmetric.(7) the field intensity obtained by gauss theorem is only excited by the charge o

6、f the gaussian surface.(1) no net charge; (2) x; (3) x; (4) square root; (5) x; (6) x; (7) x.1.4 the field intensity of a uniformly charged spherical surface is equal to the field intensity that all charges are concentrated in the center of the ballReally?Answer: in the absence of field, the ball is

7、 correct.1.5 in the attached figure, A and B are two uniform point electric bodies. S is the sphere with A concentric.(1) is the flux of S surface related to the position and charge of B?(2) does the strength of an electric field at some point on the S surface have to do with the position of B and t

8、he charge?(3) can we use gausss theorem to solve a point on S surface? Why?(1) it has nothing to do with (2) it cant (a conductor ball) and can (a medium ball).When the principle of field superposition is applied to a problem with a conductor, it is important to note that when a charged conductor ex

9、ists alone, there is a charge distributionChange the pageThey produce an electric field; When n live conductors are put together, because of electrostatic induction, the charge distribution on the conductor changes, and then,Applying the principle of superposition, the charge distribution of each co

10、nductor should be frozen, and then the charge distribution of frozen will be stored separatelyThe resulting electric field is superimposed.The force of R in radius R is to dig the ball of radius R in the point of the electric ball, and the two kinds of dredging of the attached figure (a) and (b) can

11、 be determined by gaussHow does the principle of principle and superposition work?Answer: (a) can, superposition method (compensation method); (b) noThe S1, S2, S3 and S4 in the attached figure are all the surfaces of the closed curve L as the border line (the direction of the surface normal line is

12、 shown). Have been S1E flux forone, surface S2, S3 and S4 E flux2,3 and4.The point E = 0 constant, always outside204qER PI epsilon.= constant,The point that is swept by the surface of the balloon, the E changes, from2004qER PI epsilon.= .1.8 in the attached figure, S1 and S2 are four closed surfaces

13、, with E1, E2, and E3 respectively representing the electrostatic field strongly stimulated by q1, q2 and q3, and the trial judgmentThe right and wrong of the following equations(1)oneoneone0s.qEdsEpsilon.? = ?(2)2330s.qEdsEpsilon.? = ?(3)one2320(a)s.qEEdsEpsilon.+? = ? ?(4)one12120(a)(a)s.qqEEdsEps

14、ilon.+? = ? ?(5)2321230(a)(a)s.qqEEEdsEpsilon.+ +? = ? ?(6)one1321230(a)(a)s.: QQQEEEdsEpsilon.+ + +? = ? ?Answer: (1) x; (2) x; (3) x; (4) x. (5) square root; (6) x;1.9 the electric field diagram of the two infinite uniformly charged plane with equivalent number and equivalent number is drawn.The a

15、nswer:1.10 electric field is the trajectory of electric charge in electric field? (set up thisChange the pageThe point charge is not subject to any other force except the electric field.Answer: not usually. FqE = ; FMa = ;va.t=?; Only in the uniform electric field, the static point charge motionThe

16、trace of the power line.1.11 is the following statement true? If incorrect, please give a counter-example.(1) the potential in the field of the field is also equal.(2) if two electric potentials are equal, their field strength is equal.(3) if A point field is stronger than B, then A point is bound t

17、o be higher than that of B.(4) the electric potential must be zero at zero.(5) zero field intensity must be zero.Answer: (1) incorrect.uEnn?=?A uniform electric field, for example.(2) incorrect.(3) incorrect. E big, the rate of change of electric potential is big, not necessarily U big(4) incorrect.

18、 E = 0,Un?= 0, not U must be 0, in the same amount of the same point charge.(5) incorrect. U = 0 is notUn?Must be 0, for example: at the middle point of the isoelectric point charge line.1.12 the two radii are R1 and R2 = 2R1 concentric uniform charged sphere, and the inner sphere has charge q10. Wh

19、en the outer sphere has charge q2What condition is it that the inner sphere is positive? What condition does the inner sphere have zero potential? What condition does the inner sphere have to be negative?(reference points are not far away.)The answer:12one0101442qqUThe RR PI epsilon PI epsilon.= +Or

20、:21121121121122220044The RRRRRR: QQQUEdrEdrdrdrThe rr PI epsilon PI epsilon.upUp += = + + To make the10U,21 () 02qQ +, that is,212Qq?To make the10U =,21 () 02qQ + =, that is,212Qq =?To make the10U,21 () 02qQ +, that is,212qq1.13 the field strength of the area in which the test case is necessary is z

21、ero everywhere.Proof: if E is 0, 0baba.UEdl = = , that is,abUU =, the allelic region.If its an allele, U equals 0, thats 0UEn?= =?It is.1.14 the large circle section S of the uniform charged hemisphere surface (see attached figure) is the isopotential surface. (hint: make up the other half of the sp

22、here for symmetryChange the pageThe field strength of each sphere on S is perpendicular to S.Proof: there is a field force parallel to the component on the s surface, and the rest of the ball after the other half of the ballIt should be zero, so you cant have a strong parallel component on the s sur

23、face, and its only going to be the vertical component of the fieldS surface should be equal potential surface.1.2.1 there are two point charges in a vacuum, one of which is four times the magnitude of the other. They are separated by2 -5.010 times m repulsive force1.6 N. Q:(1) how much are their cha

24、rges?(2) how much repulsive force do they have when they are 0.1m apart?Solution: set a chargeoneQ,214Qq =, by the formula1220one4qqFR PI epsilon.= can be obtained:2one220411.64 (510).qPI epsilon.?=xSolution of:oneQ =60.3310? x2,14Qq = =61.3310? x when r = 0.1, by repulsion is:1220one4 (0.1)qqFPI ep

25、silon.= = 0.4 (N)1.2.2 the charge sum of the charge of the two charges of the same sex point is Q, and the charge of each charge is as long as the distance between them is certainThe maximum interaction?Solution: if one of these charges is q, then the other charge is q-q,By the coulomb force2(a) : q

26、QqFk1 - fkr?= knowWhen 0dfdq= that is:220K.Qqr? =So the power of the two is: 2Qq =one2QQ =1.2.3 when the charged body of a point with an L is 2q and q, where the charged body of the third point is placed, it will be fittedForce is zero?Solution: set a qQ for r, then q2 q (Lr)? If the force is equal

27、to 0, the force balance condition is equal to:222(a)QQQQkkrLr =?Change the pageGet: (21) rL =?1.2.4 in rectangular coordinate system (0m, 0.1 m) and (0m, -0.1 m); Two positions have charge q = 10- 10 cThe point ofThe charged body is placed on the (0.2m, 0m) position with a charge of Q = 10- 8 cThe p

28、oint charged body, the magnitude and direction of the force of Q.According to the diagram, it can be obtained:722220.2(23.2210)0.2QqfkN?= = x+i.Direction: horizontal to right1.2.5 at the vertex of the square, each one has a charged body with a charge of q.(1) the force applied to any point

29、charge in a square center is zero.(2) if you put a charge Q in the center,So the relationship between Q and Q is zero for each charge on the vertex.Answer: (1) the symmetry is known: O point E = 0, and the force charged at O point for any charge is 0.(2) solution: set O point of charge Q, according

30、to the right diagram, 1220one4qQffA PI epsilon.= =(a)2320one42qfa.PI epsilon.=20one4one22qQFa.PI epsilon.=?If the net force of q is 0, then we have:220222000121cos454442one22qQqqaaa.PI epsilon PI epsilon PI epsilon.= +?Solution of:12(a)42Qq = + (Q should be negative point charge)1.2.6 the same sex p

31、oint charge with the same magnitude is 2a, and a pilot charge q0 is placed on the central vertical surface of the two lines, and q0 is the most stressedThe trajectory of a large point.Solution: 3222222222(a)(a)(a)QQRKQQRFk1 - fkararar = =+And 0dfDr.=Change the pageThat is:3122222222332 () ()220(a)ar

32、rarrKQQar?+? X +?=?+?i.312222222() () 0 arrar + 3? + =222Ar =So the radius of the circle is:22Ra = -The obtained curve equation is as follows:2222a.yz?+ =?. Spherical equation1.3.1 the electric field between the two belt plates with the length of 50cm and 1cm is the uniform electric field (the field

33、 strength is straight up),The initial velocity of an electron from the P point (with the upper and lower plates) is v0 = 107 m/sThe level is shot into the electric field (see attached diagram). If the electron is just going downIm going to leave the electric field on the side, and I want the magnitu

34、de of the uniform field. (ignoring the edge effect, it is considered that the external field is zero, and the electron transport is omittedInfluence of movement)Solution: the force of the electron in the electric field produces motion acceleration:0EEma =From the kinematics equation:2122dYat = =0XVT

35、 = (2dY =) xL =Solution of:20VMDEThe ex?=?1.3.2 a small ball with a fine line of 0.2 g, placed in a parallel-plate between two vertical plates (seeSet the ball with charge9 -610 x C to make the fine line of the hanging ball and the electric field Angle 600,Take the field between the two boards.Solut

36、ion: shown in the figure:00Cos30cos60eEmg =Among them:0Cos60EqT =0Cos30mgT =Solution of:030mgEtge=1.3.3 an electron ejaculation intensity is3510 times N/C, the uniform electric field that were going to enjoy in the vertical direction, the initial velocity of the electron is zero710 m/sAnd levelThe A

37、ngle of incidence is 300(see attached figure) without considering the influence of gravity, please:Change the page(1) the maximum height of the electron ascent.(2) the horizontal range of the electron back to its original height.Solution: (1) electronic force: fmaeE = =020onesin302Yvtat =? (1)00 sin

38、30yvvThe at =?When maximum height: 0yv=the000Sin30vat =?00 sin30vta.= (2)(1) the maximum height2000000sin30sin301sin302V. v.yvaaa?= x?220011Sin30 ()2vaa=?2200oneSin30 ()2va.=22002onesin30EemV =208The mvEe=(2) when returning to horizontal position:Y = 0, namely:020onesin3002eEvtm? =Solution of:002sin

39、30mvteE=Therefore, the horizontal range of the original height is:20003cos302The mvXVTeE= =1.3.4 the charge of the electron is first measured by the oil drop experiment by milligan, and the experimental apparatus designed by milligan is shown in the attached figure.A small charged oil droplet in the

40、 electric field E, regulates E to balance the electric field force on the droplet with the weight of the oil droplets, if the oil dropsThe radius of4 -1.6410 times cm, equilibrium time E is equal to51.9210 x N/C. O:Change the page(1) the oil density is 0.851 g/cm3And the absolute value of the warm d

41、rop.(2) how many times does the value of the value of the element e?Solution: (1) slightly(2) mgqE =31948.02103mgRgqEEPI rho?= = = x coulomb1.3.5 two point charges q1 = 4.0 uc and q2 = 8.0 uc are 10cm apart, and each of them is a field E of 10cm.Solution:2222021122242 cos60KqKqqqEkaaa?= + +?Input da

42、ta:60129.5210101 qqNEC= xIts the Angle from the wire to the end.1.3.6 the radius of the uniform electric ring in the attached figure is R, the total charge is q, please:(1) the axis is the field E of x.(2) where is the largest field on the axis? What is its value?(3) draw the E - x curve.Solution: (

43、1)20one4dqdER PI epsilon.=2qDQDLR PI.=222RxR = +Symmetry analysis: 0EAn =(a)32220onecos42QXDLEExdExdERxRAlpha.PI epsilon PI.= = = =+ (a)3220220one42RqxdlRxRPI.PI epsilon PI.=+(a)32220one4qxxRPI epsilon.=+(2) the curve diagram is right:(3) order 0xdEdx=That is:() ()(a)31222222322032204xRxxxRqxRPI eps

44、ilon.?+? +?=?+?Change the pageSolution: 0.72RXR = =32202one242qRERRPI epsilon.=?+?1.3.7 the charge is distributed evenly over the linear density eta in the line segment of length L.(1) a field of a point with a wire which is the point of R.(2) when L goes to infinity, the field of this point is stro

45、ng202ERetaPI epsilon.=;(3) the results obtained in RL are consistent with that of the point charge field.Solution: (1)220one4NDXdEXR PI epsilon.=+Directions:By symmetry analysis, the forces components cancel each other out.2222200onesin24LndxREdExRxRAlpha.PI epsilon.= =?+ =232220024(a)LnRdxxRPI epsilon.+ 20244nLLThe RRPI epsilon.=?+(2) when L goes to infinity:200242one4nLnERRRLLPI epsilon PI epsilon.=? =+(3) when RL? When:22200022444