Corrigé - Alain Le Rille

◦
a+q ρ =3q
4π a3r < a
ρ= 0 r > a
ρ(r)
~
E(M, ~ur, ~uθ) (M, ~ur, ~uϕ)~ E=Er(r)~ur
r
ZZ ~
E.−−→ d2Σ=4π r2Er(r) = Qint
ε0
~
E(r > a) = q
4π ε0r2~ur
Qint =ZZZ ρ(r) d3τ=4
3π r33q 4π a3=q r3 a3
~
E(r < a) = q r
4π ε0a3~ur
M−e~ F=−e~ E(r <
a)~ F=−e q r
4π ε0a3~ur~
F=−k r~urk > 0
k
~
E=−−−→ gradV
q
4π ε0r2=−dV dr⇒V=q 4π ε0r+C
V(r > a) = q
4π ε0r
q r
4π ε0a3=−dV dr⇒V=−q r2 8π ε0a3+C0
V r =a
q
4π ε0a=−q 8π ε0a+C0⇒C0=3q 8π ε0a
V(r < a) = q
8π ε0a3−r2 a2−e.V
(Ep(r < a) = −e q
8π ε0a3−r2 a2
Ep(r > a) = −e q
4π ε0r
r
a
Ep
P C P C∗
1P C∗ 2◦

Ei
Ep(r= 0) = −3e2
8π ε0aEp(r→ ∞) = 0 Ei=3e2 8π ε0a
a=3e2
8π ε0Ei
=3×1,6×10−192
8π×8,85 ×10−12 ×13,6×1,6×10−19
a= 1,6×10−10 m
α
40 nm
10−4m= 100µm
~
F=−2e q r
4π ε0a3~ur
α q =Z e
Fmax =2Z e2
4π ε0a2=2×79 ×1,6×10−192
4π×8,84 ×10−12 ×(1 ×10−10)2
Fmax = 3,6×10−6N
α
α
Mα~a =Fmax~uy⇒(Mαd2x
dt2= 0
Mαd2y
dt2=Fmax ⇒dx
dt=v0
dy
dt=Fmax Mαt⇒x=v0t
y=1
2
Fmax
Mαt2
2a=v0tf
y=1
2
Fmax
Mα
t2
f=2a2
v2
0
Fmax
Mα
=2a2
v2
0
Fmax
Mα
D
tan D=y
x=a Fmax Mαv2
0≈D
D≈10−10 ×3,6×10−6
(1,6×107)2×4×1,7×10−27 D≈10−4rad
10−2rad
α
P C P C∗
1P C∗ 2◦

α
d~σO
dt=−−→
OM ∧~ F=r ~ur∧2.Z.e2
4.π.ε0.r2~ur=~
0
α
C=k~σOk
Mα
=r ~ur∧˙r ~ur+r˙ θ ~uθ⇒
C=r2|˙ θ|
C=k~σOk
m=|(xi~ux+b ~uy)∧(v0~ux)| ⇒
C=b v0
1/r2
α
V
~
F=2Z e2
4π ε0r2~ur=−−−→
grad(Ep)⇒dEp=−2Z e2
4π ε0
dr
r2
Z0
Ep(r)
dEp=−2Z e2
4π ε0Z∞
r
dr
r2⇒Ep(r) = Z e2 2π ε0r
Ec=1
2m v2=1 2m˙r2+1 2m r2˙
θ2
Em=Ec+Ep=1
2m˙r2+1 2m r2˙
θ2+Z e2
2π ε0r
C=r2|˙ θ|Em=1
2m˙r2+m C2 2r2+Z e2 2π ε0rEm=1 2m˙r2+Epef f (r)
Epef f (r) = m C2
2r2+Z e2 2π ε0r
rmin α˙r= 0
Em=Epef f (rmin) = m C2
2r2
min
+Z e2
2π ε0rmin
=1
2m v2
0
m v2
0r2
min −Z e2
π ε0rmin −m C2= 0
rmin =
Z e2
4π ε0
1
2m v2
0
+v
u
u
t Z e2
4π ε0
1
2m v2
0!2
+b2
P C P C∗
1P C∗ 2◦

rmin =r Ep(r)
2Ec
+sr Ep(r)
2Ec2
+b2
EpEcrmin ≈b
D≈0
b r =rmin
D= 180◦
•Ep
Ec
•α◦ ◦
b
m~a =−e2
4π ε0r2~ur=−mv2 r~ur
v=e
√4π m ε0r
v=e
√4π m ε0r
2π r
T⇒e2 4π m ε0r
4π2r2
T2
r3
T2=cste
~
F=−e2
4.π.ε0.r2~ur=−−−→
grad(Ep)⇒dEp=e2
4.π.ε0
dr
r2⇒ZEp(r)
0
dEp=e2
4π ε0Zr
∞
dr
r2⇒
Ep(r) = −e2
4π ε0.r
Ec=1
2m v2=1 2me2 4π m ε0r⇒
Ec=1
2
e2
4π ε0r
Ep=−2EcEm=−Ec=Ep
2
P C P C∗
1P C∗ 2◦

h c
λ=En−Em=RHh c 1 n2−1 m2
E=−E0/n2E0=RHh c
E0= 1,1×107×6,63 ×10−34 ×3,00 ×108= 2,2×10−18 J = 2,2×10−18
1,6×10−19 eV
E0= 14 eV
Em=−E0
n2=Ep 2=−e2 8π ε0.r ⇒
r=e2
8π ε0.E0n2
n= 1 r=(1,6×10−19 )2
8π×8,84×10−12 ×2,2×10−18 r= 5,2×10−11 m
Em=−E0
n2=−Ec=1 2m v2⇒
v=q2E0
m n2
v≤q2E0
m=q2×2,2×10−18 9,1×10−31 v≤2,2×106m·s−1c
O
~
L=r ~ur∧m˙r ~ur+r˙ θ ~uθ=m r v~uz
r=e2
8π ε0.E0n2v=q2E0 m n2
~
L=me2
8π ε0.E0r2E0 mn ~uz=e2 8π ε0r2m E0
n ~uz
~
L=k
2πn ~uz
k=e2
4ε0r2m E0
=1,6×10−192
4×8,84 ×10−12 s2×9,1×10−31 2,2×10−18
P C P C∗
1P C∗ 2◦