电磁场与电磁波(双语)样卷

Beijing Jiaotong University T erminal Examination (Paper A)(the second term, 2011-2012 academic year)Course Name: electromagnetic fields and waves Teacher: Wei Y an Class__________________Student ID____________Name___________--------------------------------------------------------------------------------------------------------------------Formula of operations on scalar and vector fields in cylindrical coordinates and spherical coordinates111sin z r f ff f f ff zrr r ρφθφρρφθθφ∂∂∂∂∂∂∇=++=++∂∂∂∂∂∂a a a a a a222222222222211111()()(sin )sin sin f ff f f ff rzr rrr r ρθρρρρφθθθθφ∂∂∂∂∂∂∂∂∂∇=++=++∂∂∂∂∂∂∂∂∂22()(sin )()11111sin sin z r A A A A A r A zrrr r ρφφθρθρρρφθθθφ∂∂∂∂∂∂∇∙=++=++∂∂∂∂∂∂A2sin 11sin sin z rzrr r z r rB B B B rB r B ρφθφρφθφθρρφθθφρθ∂∂∂∂∂∂∇⨯==∂∂∂∂∂∂a a a a a a B--------------------------------------------------------------------------------------------------------------------There are 4 choices marked A, B, C, and D in every question and only one choice is correct. Please write the symbol of the choice which one you think is right in the blank in each question. In the same time, you must write down the process to obtain the solution . (5×20=100 marks)[1-3] A point charge q is enclosed in a linear, isotropic, and homogeneous dielectric medium of infinite extent. The medium has a relative permittivity εr . Suppose the point charge is located at the origin of the spherical coordinate system.1. The electric field intensity E can be expressed as (A)24πr q ra(B)204πεrq ra (C)20r 4πεεr q ra(D)20r 4πεεrq r-aY ou select ( ). Solution:2. The polarization vector P can be expressed as (A)r 20(ε1)4πεrq r -a (B)r 2r (ε1)4πεr q r -a(C)r 2r (ε1)4πεr q r--a(D)r 20(ε1)4πεr q r--aY ou select ( ). Solution:3. Determine the total bound charge Q sb , which is on the surface of the dielectric next to the point charge q. (A)r r (ε-1)/εq -(B)r/εq - (C)r (ε1)q -(D) qY ou select ( ). Solution:[4]. The plane z=0 marks the boundary between free space and a dielectric medium with a relative permittivity of εr . The Electric field intensity next to the interface in free space isx x y y z zE E E =++E a a a .Determine the Electric field intensity on the other side of the interface. (A) x x y y z zE E E ++a a a (B) /εx r x y y z z E E E ++a a a (C)/εx x y r y z zE E E ++a a a (D)/εx x y y z r zE E E ++a a aY ou select ( ). Solution:[5]. A spherical capacitor is formed by two concentric metallic spheres with inner radius a and outer radius b , b > a . The region between the two concentric spherical shells is filled with a dielectric medium with a relative permittivity of εr . Find the capacitance of the capacitor. (A)04πεab b a- (B)0r 4πεεab b a- (C)0r 4πεεab b a+ (D)04πεab b a+Y ou select ( ). Solution:[6]. A charged semicircular ring of radius b extending from φ=0 to φ=π lies in the x-y plane and is centered at origin. If the charge distribution is ksin(φ), compute the electric field intensity at P(0,0,h). (A)223/201π24πε()2y z kbb h h b ⎛⎫-+ ⎪+⎝⎭a a (B)223/201π24πε()2y z kbb h h b ⎛⎫- ⎪+⎝⎭a a (C)223/201π22πε()2y z kbb h h b ⎛⎫-+ ⎪+⎝⎭a a(D)223/201π24πε()2y z kbb h h b ⎛⎫-- ⎪+⎝⎭a aY ou select ( ). Solution:[7-8]. Charge is uniformly distributed inside an infinite long cylinder of radius a . The volume charge density is ρv .7. Calculate the electric field intensity at all points inside and outside the cylinder.(A) 02()2ε()2εv v a a a ρρρρρρρρ⎧≤⎪⎪=⎨⎪≥⎪⎩a E a(B)2()4ε()4εv v a a a ρρρρρρρρ⎧≤⎪⎪=⎨⎪≥⎪⎩a E a(C)2εv ρρρ=E a (D)204εv aρρρ=E aY ou select ( ). Solution: 8. Takeρ→∞as the zero electric potential point. Compute the electricpotential at all points inside the cylinder. (A) ()222ln 24v vaaρρρρεε+- (B)2ln 2v aρρε(C)()224va ρρε- (D)()222ln 22v vaaρρρρεε+-Y ou select ( ). Solution:[9-10] A charged ring of radius a carries a uniform charge distribution. The linear charge density is ρl .9. The electric potential at point P (0, 0, z ) on the axis of the ring is(A)(B)(C)(D)Y ou select ( ). Solution:10. The electric field intensity at point P (0, 0, z ) on the axis of the ring is (A) 223/202()l z a za z ρε⎡⎤⎢⎥+⎣⎦a (B)223/202()l z a aa z ρε⎡⎤⎢⎥+⎣⎦a (C)223/202()l z a za z ρπε⎡⎤⎢⎥+⎣⎦a (D)223/204()l z a za z ρε⎡⎤⎢⎥+⎣⎦aY ou select ( ). Solution:[11-12] We define an electric dipole as a pair of equal charges of opposite signs that are very close together. Assume that the magnitude of each charge is q and the separation between them is d. If the charges are symmetrically placed along the z axis, and the point of observation P (r, θ, φ) is quite far away so that r>>d, as illustrated in Figure P11.Figure P1111. The electric potential at point P can be written as (A)20cos 4πεqd rθ (B)20cos 2πεqd rθ (C)0cos 4πεqd rθ (D)20sin 4πεqd rθY ou select ( ). Solution:12. Calculate the electric field intensity at point P. (A) 20[2cos sin ]4πεr qd r θθθ=+E a a(B)30[2cos sin ]4πεr qd r θθθ=+E a a(C)20[cos 2sin ]4πεr qd rθθθ=-E a a(D)30[2cos sin ]2πεr qd rθθθ=+E a aY ou select ( ).Solution:[13-20] A metallic spherical shell with inner radius a and outer radius b is shown in Figure P13. The center of the spherical shell is located at the origin of the coordinate system. There is a point charge +q at point C (0, h ,0), h<a. Assume the electrical potential of the points at infinity is zero.13. If the shell is not grounded, determine its electric potential. (A)04q bπε(B)04q aπε(C)04()q a h πε+ (D)04(-)q b a πεY ou select ( ). Solution:14. If the shell is not grounded and has an electric potential V 0, find the electric potential at the center of the shell. (A) 0V(B)0011V ()4πεqh a +-(C) 0011V ()4πεqh b+-(D) 0012V ()4πεqhb a+-+Y ou select ( ). Solution:15. If the shell is not grounded, determine the electric force which the point charge +q experienced. (A) 2220/4πε(/)yqa ha h h -a(B) 2220/4πε(/)yqh aa h h -a (C)2220/4πε(/)yqb hb h h -a (D)2220/4πε(/)yqh bb h h -aY ou select ( ). Solution:16. Now suppose another point +Q is located at point D (0, d , 0), d >b . If the shell is grounded, determine the electric potential at a point P (x , y , z ) outside the metalized shell. (A)014πε⎡⎤-(B)014πε⎡⎤-(C)01/4πεQ Qb d q⎡⎤-+(D)014πε⎡⎤Y ou select ( ). Solution:17. At problem 16, when the shell is grounded, suppose the electric potential at point P (x , y , z ) outside the metalized shell is V p . Now Assume the shell is not grounded, find the electric potential at point P. (A)p V +(B)p V(C) p V +(D)014πε⎡⎤ Y ou select ( ). Solution:18. At problem 16, when the shell is grounded, determine total induced charge on the outer surface of the shell. (A) q (B)/Qb d- (C)/Qa d - (D)/q Qb d-Y ou select ( ). Solution:induced charge on the outer surface of the shell.(A) q (B)/q Qb d--(D)/Qa d-(C)/Qb dY ou select ( ).Solution:force which the point charge +Q experienced. (A) 222201/4πε(/)yqQ Q b d d d b d ⎡⎤-⎢⎥-⎣⎦a (B)222201(/)/4πε(/)y q Qb d Q Q b dd d b d ⎡⎤+-⎢⎥-⎣⎦a (C) 222201/4πε()(/)y qQ Q b dd h d b d ⎡⎤-⎢⎥--⎣⎦a (D) 2014πεyqQdaY ou select ( ). Solution:。