Maîtriser l'électrostatique : 10 exercices corrigés (Coulomb, Gauss)

r
F=
4πε r
02
∣q q ∣
1 2
ε ≈
08,854 × 10−12
≈
4πε
0
1
8,99 × 10 N ⋅
9m ⋅ 2C−2 9,0 × 109
∣q q ∣
1 2
q =
1+2μCq = 2−3μC
q
1q 2
q
2q 1 =
4πε
0
19,0 × 10 N ⋅
9m ⋅ 2C−2

F= 9,0 × 10 ×
9
≈
0,502
2 × 10 × 3 × 10
−6 −6
0,22N
q
2q 1q 2q 1
(Ox)q =
1+1μCx= 0 q = 2−2μC
x= 1 q =
3+3μCx= 2 q 2 =
4πε
0
19,0 ×
10 N ⋅
9m ⋅ 2C−2
q
1q 2q 3
F =
1→2 9,0 × 10 ×
9
=
12
10 × 2 × 10
−6 −6
1,8 × 10 N
−2
F =
3→2 9,0 × 10 ×
9
=
12
3 × 10 × 2 × 10
−6 −6
5,4 × 10 N
−2
F =
x−1,8 × 10 +
−2 5,4 × 10 = −2 3,6 × 10−2
q
3
E= F qE q
q
Q
4πε r
02
∣Q∣Q> 0 Q<
0
Q= +5μC
M Q
−1,6 × 10−19 M =
4πε
0
19,0 × 10 N ⋅
9m ⋅ 2C−2

E= 9,0 × 10 ×
9≈
0,202
5 × 10−6
1,1 × 106(QM)
Q Q > 0
q< 0 = F qE Q
1,6 × 10 ×
−19 1,125 × 10 = 61,8 × 10−13
q =
1+1μCO(0 ; 0)
q =
2−2μCA(0 ; 1) P(1 ; 0)
=
4πε
0
19,0 × 10 N ⋅
9m ⋅ 2C−2
q
1OP = 1 9,0 × 10 ×
9
=
12
10−6
9,0 × 103O P
q
2AP =
2 9,0 × 10 ×
9
=
2
2 × 10−6
9,0 × 103P A
q <
20 − ; ( 2
1
2
1) ≈
E2(−6,36 ; 6,36) × 103
=
E + E1 ≈ E2(9,00 − 6,36 ; 6,36) × 10 =
3(2,64 ; 6,36) × 10 V/m 3
E= × 2,64 + 6,36
2 2 10 ≈
36,9 × 103
1,8 × 104
E E
E S Q
int
⋅
∮S Ed = S
ε
0
Q
int

E
M E M
E E
Q
int
O R
ρ> 0 C ⋅ m−3 M
r O r <R r >R
(OM)E(OM)
O E r r
E(r) × 4πr2
r<R ρ × πr
3
43E(r) =
3ε
0
ρr
r
r>R Q =ρ× πR
3
43E(r) = =
3ε r
02
ρR3
4πε r
02
Q
O r =R
3ε
0
ρR
ρ> 0
λ> 0
M r
λ= 10nC/m r= 5,0cm =
4πε
0
19,0 × 10 N ⋅
9m ⋅ 2C−2

M
E
r
r h
E E × 2πrh
Q =
int λh
E= =
2πε r
0
λ2 × ×
4πε
0
1
r
λ
E= 2 × 9,0 × 10 ×
9
=
0,050
10 × 10−9
3,6 × 103
V= E− Vgrad
Q V =
4πε r
0
Qq
V E =
pqV A B
W =
A→Bq(V − AV ) B
q−e
q= −4μC
=
4πε
0
19,0 ×
10 N ⋅
9m ⋅ 2C−2