♣#$♦

② ♦♦♠

!♦♠

♦" ♦&" )♥&" &♥♥ ♥ ♠" m ② 0 q ♥♦ " ♣♦♥♥ ♥ ♥ &③6♥ ")0♦ 0♦ R ♦♥ ♣0" ♥♦ ♦♥&♦0" ② "♥ 06♥ " ♦&" " ♠♥ "& ; ♥

♣♦"6♥ ;0♦ "&<♥ "♣0" ♣♦0 ♥ "&♥ R &0♠♥ " 0" " ♦&"

♦-♥

=0 ♣0 ♦♥6♥ ;0♦ " ♥"0♦ ; "0♠♦" &♦" " 0③" "♦0 ♥

" ♦&" 0" =0 "&♦ ♦♥"0♠♦" "♥& ♦♥06♥ ♣0 ♣0♦♠

0③ ♥♦0♠

~N " ♣ "00 "♥♦ "" ♦♠♣♦♥♥&" 0&♥0" ♦♠♦

~N = N cos(π/3)i + N sin(π/3)j

0③

~Fe " 0 ♣♦0 ♦& 0 "♦0 ♦& ③;0 ② "

~Fe =kq2

R2

"❮❯ ❨ ❯

♦$ &♠♠♦& $③& ♣♦$ ♦♠♣♦♥♥/ &♦$ ♦/ $ ② ♣♠♦& ♦♥5♥

6$♦

$③ ♥ ❳

∑

FX = N cos(π/3)− kq2

R2= 0 (1)

$③ ♥ ❨

∑

FY = N sin(π/3)−mg = 0 (2)

♠♦& 6

N =mg

sin(π/3)

♠♣③♥♦ &/ ①♣$&♦♥ ♥

mg

sin(π/3)cos(π/3)− kq2

R2= 0

&♣♥♦ q

q = ±R

√

mg cot(π/3)

k

!♦♠

#♥♥ ♦( ♠(( ( ♠ ♠(♠ / 0 ♥( #/2( ♥ (♠♥#♦ /# ♦

♠( ♥/♦/ (#8 ①#/♠♦ ♥/♦/ (♠♥#♦ ② ♠( (♣/♦/ ( ♣ ♠♦/

/♠♥#

♥♥#/ ♣♦(?♥ 0/♦ ♠( (♣/♦/

♥♥#/ /♥ ♣0@( ♦(♦♥( ♠( ♥ #♦/♥♦ ( ♣♥#♦

0/♦

♦-♥

( /( 2#/( (♦♥ ♠(♠♦ (♥♦ ♣♦/ ♦ 0 /③ 2#/ ♥#/ ♠( (

/♣( (D ♠( (♣/♦/ ①♣/♠♥# ♥ /③ 2#/ // ② ( ♣(♦

♦ (#♥♦ ♥ 0/♦ ♥♦ ♠( /③( (♦♥ ( ♦♥(/♥♦ (#♦

#♥♠♦(

mg = kq2

x02

♦♥ x0 ♣♦(?♥ 0/♦

=⇒ x0 = q ·√

k

mg(1)

♦♥(/♠♦( ♥ ♦♦/♥ ① ♥ #♦/♥♦ ♣♦(?♥ 0/♦ x0 ♥♠♦( ♣♦/

② #?♥

F = md2x

dt2= −mg +

kq2

(x0 + x)2

K/ ♦(♦♥( ♣0@( |x| << |x0| =⇒ | xx0| << 1 ♣♦/ ♦ 0

F = md2x

dt2= −mg +

kq2

(x0 + x)2= −mg +

kq2

x02· (1 +

x

x0)−2 ∼= −mg +

kq2

x02· (1− 2

x

x0)

K/♦ #♥♠♦( 0 mg = kq2

x02

=⇒ md2x

dt2∼= −2kq2x

x03

=⇒ d2x

dt2+

2kq2

mx02· x

x0= 0

K/♦ mx02 = kq2

g

=⇒ d2x

dt2+

2g

x0· x = 0

K♦/ #♥#♦ #♥♠♦( 0 /♥ ♦(♦♥( ♣0@( (

w =

√

2g

x0

"❮❯ ❨ ❯

!♦♠

♥ ♦% '()% ♥ )(/♥♦ 1/)(♦ ♦ ② )(% (% ♥)% 1 % ♣♦♥

♥ ( ♥ ♥)(♦ ( )(/♥♦ ♥♥)( ) 1 %%)♠ %)' ♥

1(♦ ( (

♦-♥ (♠♦% ♥♦♥)(( ) 1 (③ ♥ ♦% '()% % (♦ ♦♠♠♦% x

(D♥ ( ③1( ♥(♦( ( ♥(♦( ② y ♣♥)♦ ♠♦ ♦

♥(♦( ( '() %♣(♦( %)D♥ % ♥/♦ ♥ ♦% )(% '()% ♥)('♠♦♥♦%

♥ ( ( ♥(♦( ♥♠♦% 1 % (③% ')(% 1 ①♣(♠♥) %♦♥

~F1 = kq2

L2· (cos(π

3)x− sen(

π

3)y) = k

q2

L2· ( x

2−√

3

2y)

~F2 = kq2

L2x

(③ (%)♥) %

~F = ~F1 + ~F2 =3

2k

q2

L2x−

√3

2k

q2

L2y =

√3 · k q2

L2· (√

3

2x− y

2) =

√3 · k q2

L2· (cos(π

6)x− sen(

π

6)y)

%♥)♦ (③ ♠♦% 1 %( ♣♦%) ♦( ♦♠)(I ♠♥) %)♦

% )♦( ♣♦♠♦% ( 1 (③ ♥)( ② 1 )♥ (D♥ π − π6 (%♣)♦ x

♣♦( ♦ 1 (③ %♦( 1 ♦ % ♦)(% (% ♥)% % ♥)♣( ♦♥ (③

')( %♦( 1 ♦ % ♥)(D♥ ♦♥ %I ♣( 1 1 %)' ♥ 1(♦ ♥♦% %)

( ♦% ♠D♦% % (③%

%)♥ ♥)( 1 ② ♣♦♠♦% ♥♦♥)(( %♥♦ ② ♦% %♥♦% %♦( )(/♥♦

1 1 % % )(/♥♦ 1/)(♦ %I

r

sen(π6 )

=L

sen(2π3 )

=⇒ r =L√3

K♦( )♥)♦ ( ♦% ♠D♦% % (③% ♦)♥♠♦%

√3 · k q2

L2= k

qQ

( L√

3)2

=⇒ Q =1√3· q

!♦♠

♦# &# ♣♥+# ♣♦#+# #+1♥ #♣&# ♣♦& ♥ #+♥ 2a 4♦& ♣♥+♦ ♠♦

#♠♥+♦ 6 # ♥ # +&③ ♥ ♣♥♦ ♣&♣♥& ♠#♠♦ & ♦# ♣♥+♦# ♥ 6

&③ #♦& ♥ & ♣& 6 ♥ ♣♥♦ # ♠1①♠ # ♣♦& #♠+&< ♥ &♥&♥

♥♥+& &♦ &♥&♥

♦-♥

#♠+&< ♣&♦♠ ♥+♠♥♦ #♠♦# 6 &③ #♦& & ♣& #+1

#♦& ♣♥♦ + ♣♥♦ ♣♥♦ ①② ♦♦&♥# &+#♥# ② ♣♦♥♠♦# ♠# &#

#♦& ③ #+ ♦&♠ +♥♠♦# 6 ~r = R · (cos(θ)x + sen(θ)y) ~r1 = az ② ~r2 = −az♦♥ ~r # ♣♦#B♥ & ♣& ② ~r1, ~r2 #♦♥ # ♣♦#♦♥# # &# ♣♦#+#

♥+# ♠♥♦♥# ♠1# |~r− ~r1| = |~r− ~r2| =√

R2 + a2 &③ E+& #♦& &

♣& #

~F = ~F1 + ~F2 =kqQ

(R2 + a2)3

2

· (Rcos(θ)x + Rsen(θ)y − az + Rcos(θ)x + Rsen(θ)y + az)

=2kqQR

(R2 + a2)3

2

· (cos(θ)x + sen(θ)y) .

#+♦ ♠♦# 6 &③ #♦& & ♣& #+1 #♦& ♣♥♦ &♠♥+ ♦ 6

♠♦# & # ♠①♠③& ♠B♦ #+ &③ ② ♥♦♥+&& ♦♥ ♦ & ♦♠E+&♦

#&+♦ ♣♦& ♦# ♣♥+♦# ♦♥ ♠B♦ #+ &③ # ♠1①♠ # & ♠♦# ♠①♠③&

♥B♥

|~F | = 2kqQR

(R2 + a2)3

2

= 2kqQ · f(R)

=⇒ df(R)

dR=

(R2 + a2)3

2 − 3R2(R2 + a2)1

2

(R2 + a2)3= 0

=⇒ (R2 + a2)1

2 · ((R2 + a2)− 3R2)) = 0 =⇒ R =1√2· a

❱♠♦# 6 & ♦♠E+&♦ # ♥ &♥&♥ ② # &♦ # R = 1√

2· a

"❮❯ ❨ ❯

!♦♠

❯♥ &( ♣+&, ( ,♥ ♥ ♠& ♠ ② & ♥ ♦ &♥, (♦ ♥ ♥

(7♥ ♦♥ ①&, ♥ ♠♣♦ 9,(♦

~E ( ( ♦, ♣(♠♥ ♥ A(♦ ♥

♥ +♥♦ θ ♥,( (, ② ♦

( ( A ♦,

♣♦♥ A ♦, ♣( ( ♥ ,& α(C · seg−1) ♦

♥( A ♣(♦ &, &( ♣( θ ≪ 1

♦-♥

♦ ♣(♠(♦ A & ( & ,(♠♥( ,♦& & (③& A &,+♥ ,♥♦ &♦(

♦, ② ♦ ,③( ♦♥7♥ A(♦ &♥, ( ♠&,( & (③& A

&,+♥ ,♥♦

♦♥

~FE & (③ ♣(♦ ♣♦( ♠♣♦ 9,(♦ &♦( ( q ② & ♦♠♦

~FE = q ~E

♦♠♦

~E & ♣(♦ X & ,♥(+ A

~E = Ei H♦( ♦ ,♥,♦

~FE = qEi

&(♠♦& ,♥&7♥ ,③♥♦ && ♦♠♣♦♥♥,& (,♥(&

~T = −T sin(θ)i + T cos(θ)j

♦$ & ♣ &♠$ & $③& ♣♦$ ♦♠♣♦♥♥1 ② ♣$ ♦♥4♥ 5$♦

$③ ♥ ❳

∑

FX = qE − T sin(θ) = 0 (1)

$③ ♥ ❨

∑

FY = T cos(θ)−mg = 0 (2)

T =mg

cos(θ)

♠♣③♥♦ ♥

qE − mg

cos(θ)sin(θ) = 0

&1♦ $&1 5

q =mg tan(θ)

E

$&1♦ ♥1$♦$ ♠♦& 1♥♠♦& 5 q = mg tan(θ)E ♦♠♦ θ ≪ 1 1♥♠♦& 5 tan(θ) ≈

θ D♦$ ♦ 1♥1♦ ♣♦♠♦& &$$ $ ♥ ♥4♥ 1♠♣♦ ♦♠♦

q(t) =mgθ(t)

E

$♠♦& ♦♥ $&♣1♦ 1♠♣♦ ♦1♥♠♦&

dq

dt=

mg

E

dθ

dt

♦ ♦♠♦

dqdt = α &♣♥♦ dθ

dt $&1

dθ

dt=

αE

mg

"❮❯ ❨ ❯

!♦♠

♦$ '$ +Q $ ♠♥,♥♥ $ ♥ $,♥ d $♣'3♥ ❯♥ ♣',6 '

♥, −q ② ♠$ m $ $,: ♥ ♥,'♦ $ ② ♦ ,'$ ♥ ♣<=♦ $♣③♠♥,♦

♣'♣♥' 6♥ < $ ♥ $ ♥ ' ♠$,' < ♣',6 $'

♥ ♠♦♠♥,♦ '♠3♥♦ $♠♣ ② ♥♥,' $ ♣'♦♦ ♦$3♥

♦-♥

@' '$♦' ♣'♦♠ ,③' $♥, $♣♦$3♥ $♣

♦ ♣'♠'♦ < ♠♦$ $ ' '③ '$,♥, $♦' ' −q ♦,' < $'

$ '$ $ $♥♦ ♦♣$,♦ ' ♣' $ '③$ $'F♥ ,'②♥,$ ② $'F $,♦

♦ < ♣'♦'F ♦$3♥

♦

~F1

' < ♣'♦ '③

~F1 $ ♥♥,' ♥ ♣♦$3♥ ~r1 = d/2j ② ♣',6

♣' ♥ xi @♦' ♥3♥ '③ ♥,' ♠$ '$ $

~F1 =k(−q)Q(xi + d/2j)

l3

♦

~F2

♥ $, $♦ ' < ♣'♦

~F2 $ ♥♥,' ♥ ♣♦$3♥ ~r2 = −d/2j ♣♦' ♦ ,♥,♦

'③ ,'3♥ ♥,' $, ' ② ' ♣' $'F

~F2 =k(−q)Q(xi− d/2j)

l3

$, ♠♥' '③ '$,♥, $♦' ' −q $'F

~F = ~F1 + ~F2 =−2kqQx

l3i (1)

❱♠♦& ' ♣♦♠♦& +♦♥+ +♦ ℓ ♦♥ & &2♥ d/2 ♥ ♥4♥ 5♥♦ θ ♥♦

♥ + ♥2+♦+ ♣♦+ +4♥ 2+♦♥♦♠72+

d/2

ℓ= cos(θ)

=⇒ ℓ = (d/2) cos(θ)

♦♠♦ & ♣+♦ ♥ ♣':♦ &♣③♠♥2♦ + ♣+ 5♥♦ θ & 2♠7♥

♣':♦ θ ≪ 1 ♦♥ &2♦ ♠♦& ' cos(θ) ≈ 1 ② ♣♦+ ♦ 2♥2♦ ℓ ≈ (d/2) ♥♦ ♥ &

2♥ '

~F =−16kqQx

d3(2)

D♦+ ♣+♠+ ② ♥2♦♥ & 2♥ '

~F = m~a ♥ &2 &24♥ ♣+2F & ♠

&♦+ x ② ♣♦+ ♦ 2♥2♦ m~a = md2xdt2

i &2 ♠♥+ & 2♥+5 '

−16kqQx

d3= m

d2x

dt2

=⇒ d2x

dt2+

16kqQx

md3= 0

♦♥ &2♦ +♠♦& ' + & ♥ ♠♦♠♥2♦ +♠4♥♦ &♠♣ +♥ ♥+

&+5 ♥2♦♥&

w = 4

√

kqQ

md3

♣+♦♦ ♦&4♥ &+5 ♥2♦♥&

T =2π

w=

π

2

√

md3

kqQ

"❮❯ ❨ ❯

!♦♠

♦♥%) %♥. %.)1♥ )% ♣♥.%

♥.♦ ♠♣♦ 7.)♦ ♥ ♣♥.♦ P ♥♦

) ♠♣♦ 7.)♦ ♥ :♠. ♥ ; P %. ♠② ♦ %%.♠

♦.♥

>♦) %♠.): ♠♣♦ 7.)♦ %.)? ♥ )1♥ j ♠♣♦ 7.)♦ ♣)♦♦ ♣♦)

) ♥.) %♦) ♣♥.♦ P %)? %♠♣♠♥.

~E1 =2kq

r2j

♦♠♦ % ? ) % ♦% )% .)% ♣)♦)?♥ ♥ ♠♣♦ 7.)♦ ♥ ♣♥.♦ P ② %

♦♠♣♦♥♥.% ♥ x % ♥)♥ ② ; .♥♥ ) ② %.?♥ %.♥ ♣♥.♦

>♦) ♦ .♥.♦ .♥)♠♦% ; ) ♦♠♣♦♥♥. ♥ y ♠♣♦ ② %♣7% %♠)♦

♣♦) ♣)♥♣♦ %♣)♣♦%1♥ %.♦ % ♣ ) ♥ %♥. )

♦♠♣♦♥♥. ♥ y ♠♣♦ 7.)♦ %)? Ey = E cos(θ) ♠♦♦ ♠♣♦ 7.)♦

E % ♣♦) ♥1♥

E =kq

d2 + r2

♠&' ' & . / cos(θ) = r√

d2+r2 '3 ♠♥. 3♥♠♦' /

Ey = (kq

d2 + r2· r√

d2 + r2)j =

kqr

(d2 + r2)3/2

♦ ♦ / . ' ♥3 ♣♦♠♦' '.. 3♦.♠♥3 '3 ♠♣♦ 3.♦

♦♠♦

~Ey = − kqr

(d2 + r2)3/2j

'3 ♠♣♦ '.& ♣.♦♦ ♣♦. . 3. <♦. ♦ 3♥3♦ ♠♣♦ =3.♦ .'3♥3

♥ ♣♥3♦ P '.& ♣♦. ♣.♥♣♦ '♣.♣♦'>♥

~E = ~Ey + ~Ey + ~E1

♠♣③♥♦ ♦' ♦.' ② ♦3♥♦'

~E = (2kq

r2− 2kqr

(d2 + r2)3/2)j

♥♦ P '3 ♠② ♦ ''3♠ ' 3♥ / d ≪ r ♠♣♦ =3.♦ ♥♦♥3.♦

♥ ♦ ♣♦♠♦' '.. ♦.♠ ♠' . ♦♠♦

~E = 2qk(1

r2− r

(r2 + d2)3

2

)

'3♦ ♦ ♣♦♠♦' '.. ♦.♠ /♥3 ♦♠♦

~E = 2qk(1

r2− 1

r2(1 + d2

r2 )3

2

)

=2qk

r2(1− (1 +

d2

r2)−

3

2 )

♥ ♥. ♥>♥ (1 + x)α' ♣ ♣.♦①♠. ♣♦. 3②♦. ♥♦ x ≈ 0

(1 + x)α = 1 + αx

♦♠♦ d ≪ r ' 3♥ /

dr ≈ 0 ② ♣♦. ♦ 3♥3♦ (d

r )2 ≈ 0 <♦. ♦ 3♥3♦ ♥ ①♣.'>♥

♠♣♦ ♣♦♠♦' . '♥3 ♣.♦①♠>♥

(1 +d2

r2)−

3

2 ≈ (1− 3

2

d2

r2)

♥♦ '3 .'3♦ ♥ ①♣.'>♥ ♦3♥ ♣. ♠♣♦ ' 3♥ /

~E =3qkd2

r4j

"❮❯ ❨ ❯

!♦♠

❯♥ ♣♦♦ *,-♦ ♥ ♥ ♠♣♦ *,-♦ ♥♦-♠ 2 2♣③ -♠♥, 2 ♣♦25♥

6-♦ ♦♠♦ 2 ♠2,- ♥ - ♦♥ θ 2 ♣6:♦ ♠♦♠♥,♦ ♥- ♣♦♦

2 I ♣♦♦ 2 - 2 ♣♦25♥ ♠2,- 6 2 ♦-♥,5♥ ♥- ♣-2♥,

♠♦♠♥,♦ -♠5♥♦ 2♠♣ ② ♥♥,- -♥ ♦25♥

♦-♥

2 -③2 *,-2 6 ,A♥ 2♦- 2 -2 2♦♥ 2 ♥ ♠♥, ② ♥ -5♥ ♣-♦

2,B♥ ♥ 2♥,♦2 ♦♣2,♦2 2,2 -③2 2♦♥ ♥-2 ♣♦- ♠♣♦ *,-♦

~E ② ,♥♥

♠♥, F = qE 2,♦ 2 ♣ ♣-- ♥ 2♥, -

,♦-6 2♦- ♣♦♦ 2-B ♥,♦♥2 2♠ ,♦-62 ♣-♦♦ ♣♦- -③ ♦♥

2-♥♦ -♦,5♥ 2♦- ♣♥,♦ O ♥,♦♥2 2 ,♥-B 6 ,♦-6 2♦- ♣♦♦

2

τ = −Fa sin(θ) +−Fa sin(θ) = −2Fa sin(θ) = −2qE sin(θ)

2♥♦ ♥,♦ 2 ♦,♥ ♣-,- - ♠♥♦ - ② 2 ♣ ♦♠♣-♦-

- ♦♣-♦♥ ♠♦♠♥,♦2 ♦-♠ ,♦-

♠♦2 6 ,♦-6 2♦- ♥ 22,♠ 2-B ♠♦♠♥,♦ ♥- I ♣♦- -5♥

♥- α♠♦2 -2♣,♦ 2♦- O 2 -

τ = Iα

−2qEa sin(θ) = Iα

♦♠♦ 2 ♣-♦♦ 2♦♦ ♥ ♣6:♦ 2♣③♠♥,♦ ♣♦♦ 2 ♠♣ 6 θ ≪ 1 ② ♣♦- ♦

,♥,♦ ♣♦♠♦2 - ♣-♦①♠5♥ sin(θ) ≈ θ ♠B2 ♥♦,♠♦2 6 α = d2θdt2

♠♣③♥♦

2,♦2 ♦-2 ♥ 5♥ ♥,-♦- 2 ,♥ 6

−2qEaθ = Id2θ

dt2

#$ *♥ .♥ ♣♦♠♦# #.. ♦♠♦

d2θ

dt2+

2qEaθ

I= 0

♦♥ ♦ ♠♦# 7 ♦.♥$*♥ ♥. ♣♦♦ #.9 ♥ ♠♦♠♥$♦ .♠*♥♦

#♠♣ ♦♥ .♥ ♥.

w =

√

2qEa

I

❨ ♣♦. ♦ $♥$♦ .♥ #.9

f =w

2π=

1

2π

√

2qEa

I

♣#$♦

♥"%( ♦♦♠

!♦♠

"♥♥ ♦' ♠,' ,♦ ,♦' ♦♥ ♥'' ♥' ♥♦,♠' λ1 λ2 '♣,♦'

♥ '"♥ ,③ 7 ' ,♥ ♠♦' ♠,'

♦-♥ 9♦, ② ♦♦♠ '♠♦' 7 ,③ ;", 7 ①♣,♠♥" ♥ ,

7 ♥ ♣♦'=♥ ~r ♦ ♥ '",=♥ , Ω '

~F = q ·∫

Ω

dq(~r − ~r‘)

4πǫ0|~r − ~r‘|3

9♦♥♠♦' ♠♦' ♠,' ♥ ① " ♦,♠ 7 ♥♦ , x = 0 x = L ②

♦",♦ x = L+d x = 2L+d ♥ ~r1 = x1x ~r2 = x2x ♦♥ 0 ≤ x1 ≤ L ② L+d ≤ x2 ≤ 2L+d

♦' "♦,' ♣♦'=♥ ♠,

♥♠♦' 7 ,③ 7 , '♦, ♥ ♠♥"♦ ① '

~dF21 = dq2 ·∫

Ω1

dq1(~r2 − ~r1)

4πǫ0|~r2 − ~r1|3

= dq2 ·∫ L

0

λ1dx1(x2 − x1)x

4πǫ0|x2 − x1|3

= dq2 ·λ1

4πǫ0· x ·

∫ L

0

dx1

(x2 − x1)2

= dq2 ·λ1

4πǫ0· x · 1

x2 − x1

∣

∣

∣

∣

L

0

=λ1x

4πǫ0· dq2 · (

1

x2 − L− 1

x2)

=λ1λ2x

4πǫ0· dx2 · (

1

x2 − L− 1

x2)

"❮❯ ❯

=⇒ ~F21 =λ1λ2x

4πǫ0·∫ 2L+d

L+d

(

1

x2 − L− 1

x2

)

dx2

=λ1λ2x

4πǫ0· (ln(x2 − L)− ln(x2))|2L+d

L+d

=λ1λ2x

4πǫ0· ln(

x2 − L

x2)

∣

∣

∣

∣

2L+d

L+d

=⇒ ~F21 =λ1λ2

4πǫ0· ln

(

(L + d)2

d(2L + d)

)

· x

❨ $% $ *③ ,%* / $ *♥ ♠♦$ ♠*$ ♣$ $♠♦$ / ② ♦♦♠

$%$ * ② %=♥ ♦*♠ *%

!♦♠

♠' ' -. ♦'♠♦ ♣♦' ♥ ♣'. -♠'' BCD '♦ R❬♠❪ ② ♣♦'

♦- '.9♥- ♦♥.- AB = 2R ❬♠❪ ② DE = R ❬♠❪ ♦- '♦- BC ② CD -.<♥ ♥

♦'♠♠♥. '♦- ♦♥ '- q ② −q '-♣.♠♥. ♠♥.'- A -♦' AB .♠B♥

- -.'② ♥♦'♠♠♥. ♥ ' q ♥. ' '♣'.'- ♦♥ ♥-

♦♥-.♥. -♦' .'③♦ DE ♣' A ♠♣♦ B.'♦ - ♥♦ ♥ ♥.'♦ O

♦♦♥

♦ A '♠♦- -'< ' ♠' ♦♠♣.♦ ♥ ♣'.- . ♦♠♦ ♠-.' '

'♠♦- ♠♣♦ .'♦ ♣'♦♦ ♣♦' - '♦♥- ♥ O ② ♦ ♦- -♣'♣♦♥'♠♦-

♠♣♦♥'♠♦- ♦♥♦♥ A ♠♣♦ .'♦ ♥ -. '♦♥ - ♥♦

!♦③♦

♠' - '.♥♦ ② '♦ 2R ② .♥ ♥ ♥ ' q ♥♦'♠♠♥. -.'

-. ♠♥' ♣♦♠♦- ♥' - ♥- ' ♦♠♦ λ1 = q2R

-. ♠♥' - .♦♠♠♦-

♥ ♣AJ ' dq ♠' -♠♦- A - ♠♣ '♦♥ dq = λ1dx ♦♥ dx -

♥ ♣AJ♦ .'♦③♦ ♠'

♠♣♦ .'♦ A ♣'♦ ♠♥.♦ dq -.' ♥ '♦♥ i ♣♦' ♦ .♥.♦ ♠♣♦

.'♦ A ♣'♦ -'<

"❮❯ ❯

d ~E1 = (kdq

x2)i = (

kλ1dx

x2)i

♥#&♥♦ * x = −3R *# x = −R

~E1 = k(λ1

∫

1

x2dx)i = (kλ1(−

1

x2

∣

∣

∣

∣

−R

−3R

))i

♦ - &*#

~E1 =2kλ1

3Ri

" &♦ ♦& "♦& )♥"♠♦& , &♣"♣♦♥"♥♦& ♠♣♦ )"♦ "&)♥) &)"

♥ "♦♥ hati &)♦ , ♥♠♥)♠♥) "♦ " &♥) "

♦♠♦ ♠♦& "♦ BC )♥ " ♣♦&) ♥ ② ♣♦" ♦ )♥)♦ ♠♥)♦ "

♥" ♥ ♠♣♦ &♥) ♠&♥)"& , "♦ CD )♥ " ♥) ② ♠♥)♦

" ♥" ♥ ♠♣♦ ♥)"♥) &) ♠♥" ♦ , ♠♦& " & &♠♣♠♥)

" & ♦♠♣♦♥♥)& ♥ x ♠♣♦ ② &♣& &♠"&

& ♥&& " ♣" "♦ & ♦)♥♥ ♠♥) &" , )♥ ♣♦&)

♥ " , ② , ② )♥♥ ♥ "♦

πR2

λ2 =2q

πR

λ3 = − 2q

πR

!♦ ♦♥&"♠♦& ♥ ♣,? " dq "♦ ♠♦♦ ♠♣♦ )"♦ ♣"♦

♦ ♣♦" &) " &"@ &♠♣♠♥)

dE =kdq

R2

♦♥ ② ♥& " λ2 ♣♦♠♦& &"" dq = λ2ds ♦♥ ds & ♥ ♣,?♦

)"♦③♦ "♦ &) )"♦③♦ "♦ ♦ ♣♦♠♦& &"" ♥ ♥♦♥ ♥♦ , ♦ "♦""

ds = Rdθ

"♦♠♦' ♥)♦♥' '++ ♠♣♦ ♦♠♦

dE2 =kλ2Rdθ

R2=

kλ2dθ

R

♦ ♦♠♣♦♥♥) ♥ x ♠♣♦ dE2 '+

dEx2 = dE2 cos(θ) =kλ2 cos(θ)dθ

R

♥)+♥♦ ' θ = 0 ') θ = π2

Ex2 =kλ2

R

∫ π

2

0cos(θ)dθ =

kλ2

R

❱)♦+♠♥)

~Ex2 =kλ2

Ri

!♦ ♦ ' + ♦♠♣♦♥♥) ♥ x ♠♣♦ ♣+♦♦ ♣♦+ ') +♦ '+<

♥)+♦+ ') ♠♥+

~Ex3 =kλ2

Ri

' ' )♥+ >

~E2 + ~E3 = 2kλ2

Ri

!♦③♦

q′ + > )♥ ') )+♦③♦ ♠+ ) > ♥ ♠♣♦ )+♦ ♣+♦♦

♣♦+ )♦♦ ♠+ ♥ O ♦♠♦ )♥ +♦ R ♥' + '+<

λ4 =q′

R

♦ ♣+♦♥♦ ♦+♠ > ♣+ )+♦③♦ AB ♠♦' >

~E4 = (kλ4

∫ 2R

R

1

x2dx)i = (

−kλ4

2R)i

♠♥♦ )♦♦' ♦' ♠♣♦' ♦' ♥♦ +♦

~E1 + ~E2 + ~E3 + ~E4 =2kλ1

3Ri +

2kλ2

Ri + (

−kλ4

2R)i = 0

"❮❯ ❯

$♣♥♦ λ4

λ4 =2

3(2λ1 + 6λ2)

♠♣③♥♦ ♦$ ♦1$ λ1 λ2 ② λ4 ♥♦$ ♥61♦1♠♥6 ♥♦♥61♠♦$ ♦1

1 $

q′ =2

3q(1 +

12

π)

♦♥$1♠♦$ π ≈ 3

q′ =10

3q

!♦♠

♠♣♦ +,-♦ ♣-♦♦ ♣♦- ♥ ♣♥♦ ♥♥,♦ ♥2 - ♥♦-♠ σ

♦-♥

♦♥2-♠♦2 ♣♥♦ ♥ 2,6♥ ♣♥♦ ①② ♦♦-♥2 -,2♥2 ♠♦2 ;

♠♣♦ +,-♦ ♥ ♦ ♣♦-

~E(~r) =

∫ ∫

Ω

dq(~r − ~r1)

4πǫ0|~r − ~r1|3

♦♥ ~r1 ,♦- ♣♦26♥ 2,-6♥ - Ω ; ♥- ♠♣♦ +,-♦ ♥ ~r

♥♠♦2 ; ~r = xx + yy + zz ~r1 = x1x + y1y ② |~r − ~r1|2 = (x− x1)2 + (y − y1)

2 + z2

=⇒ ~E(~r) =

∫ +∞

−∞

∫ +∞

−∞

σdx1dy1((x− x1)x + (y − y1)y + zz)

4πǫ0((x− x1)2 + (y − y1)2 + z2)3

2

♠♦2 ♠♦ -2 w = x − x1 v = y − y1 =⇒ x1 = x − w, y1 = y − v

♦♥♦ ,-♥2♦-♠6♥ 2

J =

∣

∣

∣

∣

(x1)w (x1)v

(y1)w (y1)v

∣

∣

∣

∣

=

∣

∣

∣

∣

−1 00 −1

∣

∣

∣

∣

= 1

=⇒ ~E(~r) =

∫ +∞

−∞

∫ +∞

−∞

σdwdv(wx + vy + zz)

4πǫ0(w2 + v2 + z2)3

2

=σ

4πǫ0

∫ +∞

−∞

∫ +∞

−∞

dwdv · (zz)

(w2 + v2 + z2)3

2

♠♥, ♠♦2 ♠♦ -2

w = rcos(θ)

v = rsen(θ)

J =

∣

∣

∣

∣

wr wθ

vr vθ

∣

∣

∣

∣

=

∣

∣

∣

∣

cos(θ) −rsen(θ)sen(θ) rcos(θ)

∣

∣

∣

∣

= r

=⇒ ~E(~r) =σ

4πǫ0

∫ 2π

0

∫ +∞

0

rdrdθ · (zz)

(r2 + z2)3

2

=σ

4πǫ0· 2πzz

∫ +∞

0

rdr

(r2 + z2)3

2

pero u = r2 + z2 → du = 2rdr

=σ

4ǫ0· zz

∫ +∞

z2

du

(u)3

2

=σ

4ǫ0· zz

(

2u−1

2

)∣

∣

∣

z2

+∞

=σ

2ǫ0· z

|z| z

=σ

2ǫ0· sign(z)z

"❮❯ ❯

=⇒ ~E(~r) =σ

2ǫ0· sign(z)z

♦$ & )$♦ )♦♥$♥ ♠0♦ ♠♣♦ 2$3♦ & ) ♥♦3♠ ♣♥♦

) + σǫ0

!♦♠ ♦♥%) ♥ ♣♥♦ ♥♥/♦ ♥% ) ♥♦)♠ σ ♦♥ ♥ ♦)♦

)) )♦

♠♣♦ 8/)♦ ♥ 9) ♣♥/♦ 9 ♣% ♣♦) ♥/)♦ ♦)♦

♣)♣♥)♠♥/ ♣♥♦

♦ )♦ % ♦♦ ♥ ♠) ♥♦ ♦♥/♦) )♦ ♦♥ ♥%

) ♥ ♥♦)♠ λ ② ② %/♥ ♠?% ♣)@①♠ ♣♥♦ % ♥♦♥/)) )③

8/) 9 ①♣)♠♥/ ♠)

♦.♥

❯%♥♦ ♣)♥♣♦ %♣)♣♦%@♥ ♣♦♠♦% ♥♦♥/)) ♦ %♦ ♦♥%)♥♦

%♠ /♦) ♠♣♦ 8/)♦ 9 ♥) ♥ ♣♥♦ ♥♥/♦ ♥% ) σ

%♦) ② ♥ %♦ )♦ ② ♥% ) −σ ♣♥♦ ②③ ♦♦)♥♦

♣♥♦ ♥ %/@♥ ② ① ♥ %/@♥

♦% /♠% ♥/)♦)% /♥♠♦% 9 ♠♣♦ 8/)♦ ♥)♦ ♣♦) ♣♥♦ %

~E1(~r) =σ

2ǫ0· sign(x)x

② ♠♣♦ 8/)♦ ♥)♦ ♣♦) %♦ %

~E2(~r) =−σ

2ǫ0· x(

sign(x)− x√R2 + x2

)

=⇒ ~E = ~E1 + ~E2 =σ

2ǫ0·(

x√R2 + x2

)

x

♥♠♦% 9 )③ 8/) 9 ①♣)♠♥/ ♥ ♠♥/♦ dx ♠) ♦

"❮❯ ❯

♣♥♦ ♦♥ ♦)♦ ,-. ♣♦) d~F = dq ~E

=⇒ ~F =

∫ d+a

d

dq ~E con dq = λdx

=

∫ d+a

d

λdxσ

2ǫ0·(

x√R2 + x2

)

x

=λσ

2ǫ0·∫ d+a

d

xdx√R2 + x2

· x

=λσ

2ǫ0·(√

R2 + x2)∣

∣

∣

d+a

d· x

=λσ

2ǫ0·(

√

R2 + (d + a)2 −√

R2 + d2)

· x

!♦♠

❯♥ $♣♥) *♠♠*$♦ ♥♦ ♦♥)♦$ $♦ a )♥ ♥ $ )♦) Q ♥♦$♠♠♥)

*)$ ♥ * *♣$ ♥)$♦$ ♥♥)$ ♠♣♦ 8)$♦ ♥ ♣♥)♦ O

♦.♥

*) ♣$♦♠ ♦ ♠♦* $*♦$ )③♥♦ ② ♦♦♠ ♣$ *)$♦♥* *♣$

* $ *) ② ♥♦* ♣$ *) *♦ =

~E(~0) =

∫

kdq(~r − ~r′)

‖~r − ~r′‖3

♦♠♦ ② ♠♦* *)♦ ♠♥)♦ $ ♦ ♣♦♠♦* *$$ ♦♠♦ dq = σdA ♦♥ σ *

♥* $ *♣$ ♥ ♥)$♦$ $♣♥) ♦♥*)♥) ♠♦* *$$

♠♥)♦ C$ dA ) ♠♥$ = ♣♦♠♦* $♦$$$ *♣$ ② ♣$ *)♦ $♠♦* ♦

*♥) ♦♥*$$ ♥ *$ *8$ ♦♥ *♥) *♣♦*E♥ ♣$ * $* θ ②

γ

$ ♥ ♣=G $E♥ C♥♦ θ ♠♦* = * $♦$$ ♥ ♣=G♦ $♦ ♦♥)

ds1 = R sin(γ)θ H♦$ ♦)$♦ ♦ $ ♥ ♣=G $E♥ C♥♦ γ * $♦$$ ♥ $♦

♦♥) ds2 = Rdγ *) ♠♥$ ♠♥)♦ C$ $♥$♥ ♦ ♣♦♠♦*

*$$ ♦♠♦

dA = ds1 · ds2 = R2 sin(γ)dγdθ

*) ♠♥$ ♣♦♠♦* *$$ ♠♥)♦ $ ♦♠♦

dq = σR2 sin(γ)dγdθ

"❮❯ ❯

"♦$ ♦%$♦ ♦ ♠♦- %$♠♥$ ♦- %♦$-

~r′ ② ~r $♠♥%

~r′ = ~0 ② ♣♦$ ♦%$♦

♦♠$♥♦ $ $♠♦- <

~r = R sin(γ) cos(θ)i + R sin(γ) sin(θ)j + R cos(γ)k

♦ - ? $ < ‖~r − ~r′‖ = R

-% ♠♥$ -♣♥♦ ♦- $♦$$♦- - $- ♣$ < $♦$$♥ $♣♥% -

%♥ <

~E(~0) =

∫ 2π

0

∫ 3π

2

π

2

kσ(−R sin(γ) cos(θ)i−R sin(γ) sin(θ)j −R cos(γ)k)

R3R2 sin(γ)dγdθ

"♦$ -♠%$? -♠♦- < ♠♣♦ A%$♦ $-%♥% -%$B ♥ $C♥ k ② ♣♦$ ♦ %♥%♦

♥%$ ♥%$♦$ ♥♦ ♦♥-$♠♦- - ♦♠♣♦♥♥%- $-%♥%-

~E(~0) =

∫ 2π

0

∫ π

π

2

kσ(− cos(γ)k) sin(γ)dγdθ = kσ

∫ 2π

0dθ

∫ π

π

2

(− cos(γ)k) sin(γ)dγ = πkσ

∫ π

π

2

(− sin(2θ)k))dγ

-% ♠♥$

~E(~0) = πkσ(1

2cos(2θ)

∣

∣

∣

∣

π

π

2

)k = πkσ1

2(cos(2π)− cos(π))k = (πkσ)k

♥♠♥% ♣$ -$$ -% $-%♦ ♥ ♥C♥ Q ♠♦- %$♠♥$ ♦$ σ

-%♦ - B -♥♦ < $ -% ♥♦$♠♠♥% -%$ ♥ -♣$ ♥%$♦$

%♥$♠♦- <

σ =Q

2πR2

-% ♠♥$

~E(~0) =kQ

2R2k

!♦♠

♥ ) ♠) AB .♥ ♦♥. 2L ② 3 ♣)♣♥) ♠) CD ♦♥.

L ♥♦ ♦3 .♥ ♠3♠ ) Q 3.) ♥♦)♠♠♥. ♠♥.

② )9♥ )③ ; ♥♦ ♦3 ) 3♦) ♦.)♦

♦.♥

3. ♣)♦♠ 3 )3 ♦)♠ ; ♣)♦♠ ♥.)♦) ♦ ♣)♠)♦ ; )♠♦3

3 ) ♠♣♦ @.)♦ ; ♣)♦ ♠) ). 3♦) ♥ ♣♥.♦ 3

3♠.)A ♦♠♦ ) Q 3. ♥♦)♠♠♥. 3.) 3 .♥)B ; 3♦) ♠♣♦

@.)♦ 3.)B ♥ )9♥ j

♠♣♦ @.)♦ ♣)♦♦ ♣♦) ♥ ) dq 3)B dE = kdqr2 ♦♥ r 3 3.♥ 3

) dq ♥ xi 3. ♣♥.♦ ♣9♥ dj r =√

x2 + d2 ♦♠♦ ♦

♠♦3 ♦ ♥.3 ) dq ♣♦♠♦3 3)) ♦♠♦ dq = λ1dx ♦♥ λ1 3 ♥3

♥ ) ② ♥♠♦3 ♦♠♦ λ1 = Q2L

3 ♠♣♦ ; ①♣)3♦ ♦♠♦

dE =kλ1dx

x2 + d2

❨ ♠♦3 ; ♠♣♦ 3.)B ♥ )9♥ j ② ; 3 ♦♠♣♦♥♥.3 ♥ x 3 ♥)♥ 3

♦♠♣♦♥♥.3 ♥ y 3 ♣♦♠♦3 3)) ♦♠♦ dEy = dE cos(θ) 3 ) ;

cos(θ) =d

√

(x2 + d2)

"❮❯ ❯

$% ♠♥)

dEy =kλ1ddx

(x2 + d2)3

2

♥%)♥♦ $ x = −L $% x = L

Ey = kλ1d

∫ L

−L

1

(x2 + d2)3

2

dx = kλ1d(x

(d2√

x2 + d2

∣

∣

∣

∣

L

−L

) =2kλ1L

d√

L2 + d2

$% ♠♥) )③ 4 ) ♠) ♦)③♦♥% $♦) ♥ ♣4: ) dq

♠) )% 4 $% ♥ %) y $)<

dF = dqE = dq2kλ1L

d√

L2 + y2

$% ♠) %♥ $%♥% ♥$ ) ② 4 $ ♥ %♥ ♠$♠ ) %♦% ♠

♥ %♥ ♠% )♦ 4 ♠) ♦)③♦♥% $% ♠♥) ♥♠♦$

λ2 =Q

L

$ dq = λ2dy ② )③ 4 ①♣)$ ♦♠♦

dF = λ2dy2kλ1L

y√

L2 + y2

♥%)♥♦ $ y = L $% y = 2L

F = 2kλ1λ2L

∫ L

−L

1

y√

L2 + y2dy

%♥ 4

∫

1

y√

L2 + y2dy =

1

Lln(

√

L2 + y2 − L

y)

F♦) ♦ %♥%♦

F = 2kλ1λ2L(1

Lln(

√

L2 + y2 − L

y)

∣

∣

∣

∣

∣

2L

L

)

$))♦♥♦

F = 2kλ1λ2 ln(

√5− 1

2(√

2− 1))

♠♣③♥♦ ♦$ ♦)$ λ1 ② λ2 ♥ ♥H♥ Q

F =kQ2

L2ln(

√5− 1

2(√

2− 1))

!♦♠

(③ ♣♦" ♥ "♦ * + (♥ ♥ ♥ ♥♥1♠♥1 ( ♥♦ b

② ♥+ ( ♥♦(♠ σ ② ♥ ♠( ♠♥1 (♦ ♦♥ ♥+ ( ♥♦(♠

λ ♣+1♦ ♥ ♠+♠♦ ♣♥♦ * ♥ ♥ +1♥ d

♦.♥

♦♠♦ (③ ♦♦♠ ♠♣ ( ② 1=♥ (♠♦+ ♦(♠ ♠>+ +♠♣

( (③ ♣♦( ♥ (♦ * + (♥ ♠♦+ ♦1♦+ +1♦ + (③ ♥♦

♦1(♦ +♦( ♦1(♦ A♦♥♠♦+ ♠( ♥ ① ② + ♣♥♦ ♦♥ +1>♥ ♠♦+ ♦1♦+

♣♥♦ ①② +♠1(D ♣(♦♠ ♠♦+ * (③ * ( ♥ +♦( ♥

♠♥1♦ ① ♠( +1> ( ♥ ±y ② + ♠+♠ ♥ *( ♣♥1♦ ♠(

A♦( 1♥1♦ ♥♦+ +1 ( (③ ♣♦( ♥ (♦ ♥ ① ♠(

♥ ~r2 = ~0 ~r1 = xx + yy ♦♥ −∞ ≤ x ≤ +∞ ② d ≤ y ≤ b + d ♣♦+=♥ ♠♥1♦ ①

♥ ① ♠( ② ♣♦+=♥ ♥ (+♣1♠♥1 +D 1♥♠♦+ * (③

+♦( 1 ♠♥1♦ +

~dF21 = dq2

∫ ∫

Ω

dq1 · −~r1

4πǫ0|~r1|3, con dq1 = σdxdy

= −dq2

∫ +∞

−∞

∫ b+d

d

σdxdy · xx

4πǫ0(x2 + y2)3

2

− dq2

∫ +∞

−∞

∫ b+d

d

σdxdy · yy

4πǫ0(x2 + y2)3

2

= −dq2σy

4πǫ0

∫ +∞

−∞

∫ b+d

d

dxdy · y(x2 + y2)

3

2

=dq2σy

4πǫ0

∫ +∞

−∞

(

1√

x2 + y2

)∣

∣

∣

∣

∣

b+d

d

· dx

=dq2σy

4πǫ0

∫ +∞

−∞

(

1√

x2 + (b + d)2− 1√

x2 + d2

)

· dx, pero dq2 = λdx2

=⇒~dF21

dx2=

λσy

4πǫ0

∫ +∞

−∞

(

1√

x2 + (b + d)2− 1√

x2 + d2

)

· dx

=λσy

2πǫ0· ln

(

x +√

x2 + (b + d)2

x +√

x2 + d2

)∣

∣

∣

∣

∣

+∞

0

=λσy

2πǫ0·

lımx→+∞

ln

1 +

√

1 + ( b+dx

)2

1 +

√

1 + ( dx)2

− ln

(

b + d

d

)

=λσy

2πǫ0· ln

(

d

b + d

)

=⇒~dF21

dx2=

λσy

2πǫ0· ln

(

d

b + d

)

"❮❯ ❯

!♦♠

♦♥%) ♥ ♠) )01♥♦ ♥♥0♦ ♥ %0♥ R ♥ ♠) %♠)) )♦

R ♦♠♦ % ♥ ) ♠♦% %07♥ ♥♦)♠♠♥0 )♦% ♦♥ ♥% ♥ )

λ ♥♥0) )③ )♣%<♥ ♥0) ♠%

♦.♥

♦ ♣)♠)♦ > ♠♦% % ) ♠♣♦ @0)♦ ♣)♦♦ ♣♦) ♥ ♠) ♥♥0♦ ♥

%0♥ d @ ♠♣♦ @0)♦ 0♥)7 )<♥ j ② > %) ♠) ♥♥0♠♥0

)♦ % ♥)♥ % ♦♠♣♦♥♥0% ♠♣♦ ♥ x %0♦ % ♣ ♣)) ♥ %♥0 )

♦♠♣)♦♠♦% %0♦ ♥♦ ♠♣♦ ♠♣♦ @0)♦ > ♣)♦ ♥ ♣>C )

dq ♥ ♣♥0♦

~r′ = xi ♠) %♦) ♣♥0♦ ~r = dj %)7

d ~E =kdq(~r − ~r′)

|~r − ~r′|3=

kdq(−xi + dj)

(x2 + d2)3

2

♦) % 0♥ > dq = λdx ② ♣♦) ♦ 0♥0♦

d ~E =kλdx(−xi + dj)

(x2 + d2)3

2

♥0)♥♦ % −∞ +∞

~E =

∫ +∞

−∞

kλdx(−xi + dj)

(x2 + d2)3

2

= kλ((−∫ +∞

−∞

x

(x2 + d2)3

2

)i + (

∫ +∞

−∞

d

(x2 + d2)3

2

)j)

♥)♥ f(x) = x

(x2+d2)3

2

+ ♠♣. ② ♣♦. ♦ 2♥2♦ ♥2. +2 ♥)♥ ♥ ♥ ♥2.♦

♣. 2♦♠ ♦. 8♦. ♦2.♦ ♦ ♥)♥ g(x) = d

(x2+d2)3

2

+ ♣. ② ♥2. +2 ♥)♥

+ −∞ +∞ +.: ♦ ♥2. ① + +∞

∫ +∞

−∞

g(x)dx = 2

∫ +∞

0g(x)dx

8♦. ♦ 2♥2♦ ①♣.+)♥ ♣. ♠♣♦ + .

~E = 2kdλ

∫ +∞

0

1

(x2 + d2)3

2

j

2♥ A

∫

1

(x2+d2)3

2

= x

d2(x2+d2)1

2

8♦. ♦ 2♥2♦

~E = 2kdλ(x

d2(x2 + d2)3

2

∣

∣

∣

∣

∣

∞

0

)j =2kλ

dj

♦. A +♠♦+ ♦♠♦ + ♠♣♦ D2.♦ ♥.♦ ♣♦. ♥ ♠. ♥♥2♦ ♣♦♠♦+

. .③ +♦. ♠. +♠.. 8. . +2 .③ ♦♥+..♠♦+ ♣AH♦+

2.♦③♦+ ♦♥ . dq ♥ .③ ♣ +♦. +.: ♣♦. ♥)♥dF = dqE

♦♥ E + ♠♣♦ D2.♦ ♣.♦♦ ♣♦. ♠. ♥♥2♦ 2. ♥ A + ♥♥2.

. dq ♦...D ♦+ ♠♥2♦+ . ♠. ♥ ♥)♥ ♥ :♥♦ θ A .♦..

♠. +♠.. ♦♥+.. +♥2 .

♠♣♦ D2.♦ ♣.♦♦ ♣♦. ♠. ♥♥2♦ ♣♥ +2♥ + +2

♦♠♦ 2. ♠♥2♦ dq . . .③ ♣.♦ +♦. ♥

+2+ .+ 2♠D♥ ♦ .: 2. ♠♥2♦ dq ♥ ♥)♥ :♥♦ θ A

①♣.+ ♦♠♦ d(θ) = R + R sin(θ) ♠:+ ♠♥2♦ . dq + ♣♦. ♥)♥

λds ♦♥ ds + ♥ ♣AH♦ 2.♦③♦ ♠. .. ♣♦♠♦+ +.. ♥ ♥)♥

:♥♦ θ ♦♠♦ ds = Rdθ +2 ♠♥. .③ +♦. ♥ . dq +.:

dF = dqE = λRdθ · 2kλ

R + R sin(θ)=

2kλ2dθ

1 + sin(θ)

♥2.♥♦ + θ = 0 θ = π ♦2♥♠♦+ .③ 2♦2 +♦. ♠. +♠.. 8.

+2♦ 2③♠♦+ ♦ A

"❮❯ ❯

∫

1

1 + sin(θ)=

sin(θ)− 1

cos(θ)

♦♥ &'♦ '♥♠♦& ) .③ .&'♥' &

F = 4kλ2