九十七學年第一學期PHYS2310電磁學 期中考試題(共兩頁) …thschang/notes/97 fall midterm II.pdf · - 1 - 九十七學年第一學期PHYS2310電磁學 期中考試題(共兩頁)

- 1 -

九十七學年第一學期 PHYS2310 電磁學 期中考試題(共兩頁)

[Griffiths Ch.1-3] 補考 2008/12/23 10:10am–12:00am, 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 formulas

1 1ˆˆ ˆsin

V V VVr r rθ θ φ

∂ ∂ ∂∇ = + +

∂ ∂ ∂r θ φ and 2

2

1 1 1( ) (sin )sin sinrr v v v

r r r rθ φθθ θ θ φ

∂ ∂ ∂∇ ⋅ = + +

∂ ∂ ∂v

1. (8%,12%) 2 2 2ˆˆ ˆcos cos cos sinr r rθ φ θ φ= + −v r θ φ

(a) Compute ∇ ⋅ v . (b) Check the divergence theorem using the volume shown in the figure (one octant of the sphere of radius R). [Hint: Make sure you include the entire surface.]

2. (10%, 10%) Suppose the potential at the surface of a hollow hemisphere is specified, as shown in

the figure, where 1( , ) 0V a θ = , 22 0( , ) (2cos 5cos sin )V b Vθ θ θ θ= − , 3 ( , 2) 0V r π = . V0 is a

constant. (a) Show the general solution in the region b r a≤ ≤ and determine the potential in the region

b r a≤ ≤ , using the boundary conditions. (b) When 2 0( , ) sinV b Vθ θ= and 1 3( , ) ( , 2) 0V a V rθ π= = , how do you solve this problem?

Please explain as detailed as possible. [Hint: 2

0 1 2( ) 1, ( ) , ( ) (3 1) / 2,P x P x x P x x= = = − and 33 ( ) (5 3 ) / 2P x x x= − .]

- 2 -

3. (7%, 7%, 6%) The potential of some configuration is given by the expression ( ) rV Ae rλ−=r , where A and λ are constants. (a) Find the energy density (energy per unit volume). (b) Find the charge density ρ(r). (c) Find the total charge Q (do it two different ways) and verify the divergence theorem.

4. (7%,7%,6%) A uniform line charge λ is placed on an infinite straight wire, a distance d above a

grounded conducting plane. (a) Find the potential V in the region above the plane. (b) Find the surface charge density σ induced on the conducting plane. (c) Find the force on the wire per unit length. [Hint: Use the method of images.]

5. (8%, 6%, 6%) Consider a hollowed charged sphere with radius R and uniform charge density ρ

as shown in the figure. The inner radius of the spherical cavity is R/2. (a) If the observer is very far from the charged sphere, find the multiple expansion of the

potential V in power of 1/r (b) Find the dipole moment p. (c) Find the electric field E up to the dipole term. [Note: Specify a vector with both magnitude and direction.]

R

R2

ρ

r

- 3 -

1. (a)

2 2 2ˆˆ ˆcos cos cos sinr r rθ φ θ φ= + −v r θ φ

22

2 2 2 22

1 1 1( ) (sin )sin sin

1 1 1( cos ) (sin cos ) ( cos sin )sin sin

cos cos4 cos cos cossin sin

4 cos

rr v v vr r r r

r r r rr r r r

r r r

r

θ φθθ θ θ φ

θ θ φ θ φθ θ θ φ

θ θθ φ φθ θ

θ

∂ ∂ ∂∇ ⋅ = + +

∂ ∂ ∂∂ ∂ ∂

= + + −∂ ∂ ∂

= + −

=

v

(b)

/ 2 / 2 / 22 3

0 0 0 0 0/ 2 / 2

4 4 4

0 0 0

The divergence theorem

(4 cos ) sin (4 cos sin )2

(4sin 2 ) sin 2 216 4 4-plane -plane -pla

V SR R

VR

S

d d

d r r drd d r drd

dr d R d R

d xy yz zx

π π π

π π

τ

πτ θ θ θ φ θ θ θ

π π πθ θ θ θ

∇ ⋅ = ⋅

∇ ⋅ = =

= = =

⋅ = + +

∫ ∫

∫ ∫ ∫ ∫ ∫ ∫

∫ ∫ ∫

∫

v v a

v

v a

2

/ 22 3 3 4

0 0

2

ne curved surface

ˆ-plane: , 0, ( cos sin ) 0, 1ˆ-plane: , 2, ( cos sin ) cos , ( cos )4

ˆ-plane: , 2, ( c

R

xy d rdrd d r rdrd

yz d rdrd d r rdrd r drd r drd R

zx d rdrd d r

π

θ φ θ φ θ

θ φ π θ φ θ θ θ θ θ

φ θ π

+

= − = ⋅ = =

= = ⋅ = − = − − = −

= = ⋅ =

∫ ∫

a φ v a

a φ v a

a θ v a/ 2

3 3 4

0 04

2 2 2

/ 2 / 2 44

0 0

4 4 4

1os ) cos , ( cos )4

ˆcurved surface: sin , , ( cos ) sin sin 2 ,2

( sin 2 )2 4

1 104 4 4

R

S

rdrd r drd r drd R

Rd R d d r R d R R d d d d

R d d R

d R R R

π

π π

φ φ φ φ φ φ

θ θ φ θ θ θ φ θ θ φ

πθ θ φ

π

= =

= = ⋅ = =

=

⋅ = − + +

∫ ∫

∫ ∫

∫

a r v a

v a 4

4 V

R dπ τ= = ∇ ⋅∫ v

2. (a)

12 3

2 0 0 0 3

3

( 1)

0

(i) ( , ) 0Boundary condition (ii) ( , ) (2cos 5cos sin ) (5cos 3cos ) 2

(ii) ( , 2) 0

General solution ( , ) ( ) (cos )

V aV b V V V PV r

V r A r B r P

θθ θ θ θ θ θθ π

θ θ∞

− +

=

= = − = − = = =

= +∑

- 4 -

( 1) 2 1

0

( 1)0 3

03 4

3 3 0

B.C. (i) ( , ) ( ) (cos ) 0

B.C. (ii) ( , ) ( ) (cos ) 2 (cos )

Comparing the coefficiency 2 , 0 for 0,1,2,4,5,...B.C. (i

V a A a B a P B A a

V b A b B b P V P

A b B b V A B

θ θ

θ θ θ

∞− + +

=∞

− +

=−

→ = + = ⇒ = −

→ = + =

⇒ + = = = =

∑

∑

3 43 3 32

4 4 70 0

3 37 7 7 7

34 3 4 7 40 0

7 7 7 7

ii) ( , ) ( ) (0) 0 0 except 3,

2 2 and

2 2 5cos 3cos( , )2

V r A r B r PA B

V b V b aA Bb a b a

V VV r b r b a rb a b a

πθ

θ θθ

−

−

→ = = + =⇒ = = =

= = −− −

− ∴ = − − −

(b)

1

2 0

3

( 1)

0

( 1) 2 1

0

(i) ( , ) 0Boundary condition (ii) ( , ) sin

(ii) ( , 2) 0

General solution ( , ) ( ) (cos )

. . (i) ( ) (cos ) 0

. . (iii)

V aV b VV r

V r A r B r P

BC A a B a P B A a

BC

θθ θθ π

θ θ

θ

∞− +

=∞

− + +

=

= = = =

= +

+ = ⇒ = −

∑

∑( 1)

02 1

010

1

1 0

( ) (0) 0 =1,3,5,... only odd terms survive

. . (ii) ( ) (cos ) sin

0 if ( ) ( ) (cos ) (cos )sin 2 , if

2 1

(

A r B r P

aBC A b P Vb

P x P x dx P P d

A b

π

θ θ

θ θ θ θ

∞− +

=+∞

+=

′ ′−

+ = ⇒

− =

′ ≠= = ′ = +

∑

∑

∫ ∫2 1

01 0 00

1

02 1 2 1 0

) (cos ) (cos )sin sin (cos )sin

2 1( ) sin (cos )sin2

But 0 for =1,3,5,... It does not make sense. Why?

Add an artifical boundary cond

a P P d V P dbbA V P d

b a

A

π π

π

θ θ θ θ θ θ θ θ

θ θ θ θ

+∞

′ ′+=

+

+ +

− =

+=

−

=

∑ ∫ ∫

∫

0

2

0

0

2

sin for 02ition ( , )

sin for 2

sin for 02or ( , )

0 for2

VV b

V

VV b

πθ θ πθ

πθ θ π

πθ θ πθ

π θ π

≤ ≤=

− ≤ ≤

≤ ≤=

≤ ≤

- 5 -

3. (a)

2 2

2 22 20 0

4

( 1)ˆ ˆ ˆ( )

( 1)Energy density2 2

r r r r

r

e re e r eV A A Ar r r r

r eE Ar

λ λ λ λ

λ

λ λ

ε ε λ

− − − −

−

∂ − − += −∇ = − = − = ∂

+= =

E r r r

(b)

0 0 0 02 2 2

3 3 32

2

2 2 2

30

ˆ ˆ( 1) ˆ( ) ( 1) ( ) (( 1) )

ˆ( ) 4 ( ) and ( 1) ( ) ( )

ˆ ˆ 1ˆ(( 1) ) (( 1) ) (( 1) )

4 ( )

rr r

r

r r r r

r eA A r e A r er r r

r er

r e r e r e er r r r r r

A

λλ λ

λ

λ λ λ λ

λρ ε ε ε λ ε λ

πδ λ δ δ

λλ λ λ

ρ ε πδ

−− −

−

− − − −

+= ∇ ⋅ = ∇ ⋅ = + ∇ ⋅ + ⋅∇ +

∇ ⋅ = + =

∂ ∂⋅∇ + = ⋅ + = + = −

∂ ∂

=

r rE r

r r r r

r r r

r2

rer

λλ − −

(c) 2 2

3 20 0

02

2 2

0 0 0 0

[4 ( ) ] 4 [1 ]

1 0

r r

v v r

r r r x

r r r x v

Q d A r e d A e r drr r

e r dr e rdr rde xde Q dr

λ λ

λ λ λ

λ λρ τ ε πδ τ πε

λ λ λ ρ τ

∞− −

=∞ ∞ ∞ ∞

− − − −

= = = =

= = − = +

= = − = − = − ⇒ = =

∫ ∫ ∫

∫ ∫ ∫ ∫ ∫

0 0 0

Use Gauss's law, the charge enclosed in a sphere of radius 4 ( 1) The total charge 4 ( 1) 0R R

R R RS

RQ d A R e Q A R eλ λε πε λ πε λ− −

→∞ =∞= ⋅ = + ⇒ = + =∫ E a

4. (a)

0

0

00

0

Assume the image line charge of is placed at a distance below the plane. 1 ˆUsing the Gauss's law, the electric field outside a line charge is .

2

So ln ( ) ( )2

r

refr

d

rrV d V r V rr

λλλ

πελπε

−

=

= − ⋅ = = −∫

E r

E l

2 20 0

2 22 2 2 20 0

( )ln ln ln2 4 ( )( ) ( )

r r x d yV V Vx d yx d y x d y

λ λπε πε+ −

+ + = + = − = − +− + + +

(b)

- 6 -

2 2

0 0 2 2 2 2 2 200

2 2 2 2

2 2

2

2 2

2

( ) 2( ) 2( )ˆ ln4 ( ) 4 ( ) ( )

4 4

Simple check:

Let tan , secsecse

xxx

x d y x d x dEx x d y x d y x d y

d dd y d y

ddy dyd y

y d dy d ddd

λ λσ ε επ π

λ λπ π

λλ σπ

θ θ θλ θλπ

==

∞ ∞

−∞ −∞

∂ + + + −= ⋅ = = − = − − ∂ − + + + − +

= − = −+ +

′ = = −+

= =

′ = −

∫ ∫

E n

/ 2

2/ 2 c

dπ

π

θ λθ−

= −∫

(c)

2

0 02 (2 ) 4

dF Edq E ddF Ed d d

λλ λλ λ

πε πε

= =

= = =

5. (a)

3 310 0 2

12

12

Consider this problem as two charge spheres, one with charge density the other with opposite charge density .

1 4 1 4( ) and ( ( ) )4 3 3 24

1 1 (1 ( )cos )

big smallRV R V

r

Rr r

ρρ

π πρ ρπε πε

θ

−

= = −−

= + +−

r R

r R…

13 3 2

0 0

3 3 3

0 0

20 0

Using the principle of superposition, we find,

1 4 1 4( ) ( ( ) )(1 ( )cos )4 3 4 3 2

1 7 4 1 4 4 ( ) ( ( ) )( ) cos , let 4 8 3 4 3 2 2 3

1 7 1 ( ) cos4 8 8 24

RRV Rr r r

R RR Q Rr r rQ Q R

r r

π πρ ρ θπε πε

π π πρ ρ θ ρπε πε

θπε πε

= − + +

= − + =

= − +

…

…

…

(b)

3

20 0

431 7 1 ( ) cos

4 8 8 24The first term is the monopole term and the second term is the dipole term.

ˆSo the dipole moment .16

Q R

Q Q RVr r

QR

πρ

θπε πε

=

= − +

= −p z

…

(c)

- 7 -

20 0

1 7 1 ( )cos4 8 8 24

1 1ˆˆsin

Q Q RVr r

V V VVr r r

θπε πε

θ θ φ

= − +

∂ ∂ ∂= −∇ = − − −

∂ ∂ ∂E r θ

…

2 3 30 0 0

ˆ

1 7 2 ˆˆ cos sin84 4 4Q p p

r r rθ θ

πε πε πε

= − −

φ

r θ