九十七學年第一學期 PHYS2310 電磁學期中考試題 ( 共兩頁 )

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1 九十七學年第一學期 PHY 電磁學期中考試題 ( 共兩頁 ) [Giffiths Ch.-] 補考 8// :am :am, 教師 : 張存續記得寫上學號, 班別及姓名等 請依題號順序每頁答一題 Useful fomulas V ˆ ˆ V V = + θ+ V φ ˆ an θ sinθ φ v = ( v) (sin ) + θvθ + v sinθ θ sinθ φ φ. (8%,%) cos ˆ cos ˆ cos sin v= θ+ φθ θ φφ ˆ Compute v. Check the ivegence theoem using the volume shown in the figue (one octant of the sphee of aius ). [Hint: Make sue you inclue the entie suface.]. (%, %) uppose the potential at the suface of a hollow hemisphee is specifie, as shown in the figue, whee V ( a, θ ) =, V( b, θ ) = V(cosθ 5cosθsin θ), V (, ) =. V is a constant. how the geneal solution in the egion b a an etemine the potential in the egion b a, using the bounay conitions. When V(, b θ ) = Vsinθ an V ( a, θ ) = V (, ) =, how o you solve this poblem? Please explain as etaile as possible. [Hint: P( x) =, P( x) = x, P( x) = (x )/, an P ( x ) = (5 x x )/.] - -

2 λ. (7%, 7%, 6%) The potential of some configuation is given by the expession V() = Ae, whee A an λ ae constants. Fin the enegy ensity (enegy pe unit volume). Fin the chage ensity ρ(). (c) Fin the total chage Q (o it two iffeent ways) an veify the ivegence theoem.. (7%,7%,6%) A unifom line chage λ is place on an infinite staight wie, a istance above a goune conucting plane. Fin the potential V in the egion above the plane. Fin the suface chage ensity σ inuce on the conucting plane. (c) Fin the foce on the wie pe unit length. [Hint: Use the metho of images.] 5. (8%, 6%, 6%) Consie a hollowe chage sphee with aius an unifom chage ensity ρ as shown in the figue. The inne aius of the spheical cavity is /. If the obseve is vey fa fom the chage sphee, fin the multiple expansion of the potential V in powe of / Fin the ipole moment p. (c) Fin the electic fiel E up to the ipole tem. [Note: pecify a vecto with both magnitue an iection.] ρ - -

3 . cos ˆ cos ˆ cos sin v= θ+ φθ θ φφ v = ( v) (sin ) + θvθ + vφ sinθ θ sinθ φ = θ + θ φ + sin sin θ φ cosθ cosθ = cosθ + cosφ cosφ sinθ sinθ = cosθ ˆ ( cos ) (sin cos ) ( cos sin ) θ θ θ φ The ivegence theoem V vτ = v a V / / / vτ = (cos θ) sin θθφ = ( cosθsin θ) θ / / = (sin θ) θ = sin θ θ = 6 v a= xy-plane + yz-plane + zx-plane + cuve suface xy a θφˆ φ v a θ φ θ -plane: =, =, = ( cos sin ) =, -plane:,, ( c / yz-plane: a = θφˆ, φ =, v a= ( cosθsin φ) θ = cos θθ, ( cos θ) θ = zx a = φθˆ θ = v a= / cuve suface: a= sin θθφˆ, =, v a= ( cos θ) sinθθφ = sin θθφ, / / ( sin θ) θφ = v a = + + = = τ v V os φ) φ = cos φφ, ( cos φ) φ =. (i) V ( a, θ ) = Bounay conition (ii) V ( b, ) V (cos 5cos sin ) V (5cos cos ) V P (ii) V (, θ = ) = θ = θ θ θ = θ θ = ( + ) Geneal solution V(, θ) = ( A + B ) P(cos θ) = - -

4 ( + ) + B.C. (i) V( a, θ) = ( Aa + Ba ) P(cos θ) = B = Aa = ( + ) B.C. (ii) Vb (, θ) = ( Ab + Bb ) P(cos θ) = VP (cos θ) = Ab + Bb = V A = B = = B.C. (iii) V(, θ = ) = ( A + B ) P() = A = B = except =, 7 Vb Vba A = an B 7 7 = 7 7 Compaing the coefficiency, fo,,,,5,... b a b a V V 7 5cos θ cosθ V(, θ ) = b ba b a b a (i) V ( a, θ ) = Bounay conition (ii) V( b, θ) = Vsin θ (ii) V (, θ = ) = ( + ) Geneal solution V(, θ) = ( A + B ) P(cos θ) = ( + ) + BC.. (i) ( Aa + Ba ) P(cos θ ) = B = Aa = ( + ) BC.. (iii) A + B P = = BC.. (ii) + a A( b ) P(cos θ) = V sinθ + = b ( ) () =,,5,... only o tems suvive if P( x) P ( x) x P(cos θ) P (cos θ)sinθθ = =, if = + + a A ( b ) P(cos θ) P (cos )sin θ θθ = V sin θp (cos θ)sinθθ + b = + b + A = ( ) V sin θp(cos θ)sinθθ + + b a But A = fo =,,5,... It oes not make sense. Why? V sin θ fo θ A an atifical bounay conition V ( b, θ ) = V sin θ fo θ V sin θ fo θ o V ( b, θ ) = fo θ - -

5 . λ λ λ λ e λe e ( λ+ ) e E= V = A ( )ˆ = A ˆ A ˆ = λ ε ε ( λ+ ) e Enegy ensity = E = A λ ( λ+ ) e λ ˆ ˆ λ ρ = ε ˆ E= εa( ) = ε A( λ+ ) e ( ) + ε A (( λ+ ) e ) ˆ λ ( ) = δ ( ) an ( λ+ ) e δ ( ) = δ ( ) ˆ λ ˆ λ λ λ λ (( λ+ ) e ) = ˆ (( λ ) e ) (( λ ) e ) e + = + = ρ = ε A δ ( ) λ e λ (c) λ λ λ λ = ρ τ = ε δ τ = ε + v v = λ λ λ λ x Q A[ ( ) e ] A[ e ] e = e λ = λe = xe = Q = ρτ = = = = x= v Use Gauss's law, the chage enclose in a sphee of aius Q A( ) e λ λ = ε E a= ε λ + The total chage Q = ε A( λ+ ) e = =. Assume the image line chage of λ is place at a istance below the plane. λ Using the Gauss's law, the electic fiel outsie a line chage λ is E= ˆ. ε λ o V = E l = ln = V( ) Vef ( ) ε λ λ ( x + ) + y V = V+ + V = ln ln = ln ε ε ( x ) + y ( x ) + y ( x+ ) + y - 5 -

6 λ ( x+ ) + y λ ( x+ ) ( x ) σ = ε ˆ En = εex = ln = x ( x ) + y ( x ) y ( x ) y x= λ λ = = + y + y x= λ imple check: λ = σy = y = θ y = Let tan, sec λ λ = (c) / sec sec / θ θ θ θ = λ θ F = Eq = Eλ F λ λ = Eλ = λ = ε ( ) ε + y y 5. Consie this poblem as two chage sphees, one with chage ensity ρ the othe with opposite chage ensity ρ. Vbig = ( ρ ) an Vsmall = ( ρ ( ) ) ε ε = ( + ( ) cos θ + ) Using the pinciple of supeposition, we fin, ( ρ ) ( ρ ( ) )( ( )cos θ ) ε ( ρ ) ( ρ ( ) )( )cos θ, let Q ρ ε V = + + ε 7 = + = ε 8 7Q Q = ( )cosθ + ε 8 ε 8 Q= ρ 7Q Q V = ( )cosθ + ε 8 ε 8 The fist tem is the monopole tem an the secon tem is the ipole tem. Q o the ipole moment p= zˆ. 6 (c) - 6 -

7 7Q Q V = ( )cosθ + ε 8 ε 8 V V ˆ V E= V = ˆ θ φˆ θ sinθ φ 7Q p p = cosθ ˆ sinθθˆ 8 ε ε ε - 7 -