פתרון בגרות · חשמל
✦ בנוי על ידי פהד גאנם ✦
العربية
☀ מצב בהיר
1
▾
✕
r = 0.9 m | m = 0.01 kg | v₀ = 2 m/s
k = 9·10⁹ N·m²/C² | e = 1.6·10⁻¹⁹ C
2.00
44.4
2.83
1.25
F
e
= (−e)·E ⟸
q
A
+ q
B
+ q
C
= 0
N = m·g
F
C
=
k·|q
A
|·|q
C
|
r²
k·q²
r²
=
m·v₀²
r
q² =
m·v₀²·r
k
=
0.01 · 2² · 0.9
9·10⁹
=
0.036
9·10⁹
= 4·10⁻¹² C²
q =
√
4·10⁻¹²
= 2·10⁻⁶ C = 2
μC
F =
9·10⁹ · (2·10⁻⁶)²
0.9²
= 4.44·10⁻² N ,
m·v₀²
r
=
0.01 · 4
0.9
= 4.44·10⁻² N
Δn =
|q|
e
=
2·10⁻⁶
1.6·10⁻¹⁹
= 1.25·10¹³
k·q²
r²
=
m·v₀²
r
⟸
2
▾
✕
a = 0.6 m | q₁ = q₂ = +40·10⁻⁹ C | k = 9·10⁹ N·m²/C²
q₃ = −40 μC | x
K
= 6 cm | a
3
= 2 m/s²
40 nC
1732
1200
2400
150
6.0
E₁ = E₂ =
k·q
a²
=
9·10⁹ · 40·10⁻⁹
0.6²
=
360
0.36
= 1000
V/m
E = 2 · E₁ · cos30° = 2 · 1000 · 0.866 = 1732
V/m
E =
√
3
· 1000 = 1.73·10³
V/m
V₁ = V₂ =
k·q
a
=
9·10⁹ · 40·10⁻⁹
0.6
=
360
0.6
= 600
V
V = V₁ + V₂ = 600 + 600 = 1200
V
E
1
= E
2
=
k·q
r²
, r =
a
2
= 0.3 m
E
tot
= E
1
− E
2
= 0 ,
V = 2 ·
k·q
a
2
= 2 ·
360
0.3
= 2400
V
E
x
= −
ΔV
Δx
E = −
6 − 0
0.02 − 0
= −
6
0.02
= −300
V/m
E = −
0
0.02
= 0
E = −
0 − 6
0.08 − 0.04
= +
6
0.04
= +150
V/m
E(K) = +150
V/m
⟸
F = q₃ · E = (−40·10⁻⁶) · (+150) = −6·10⁻³ N
|F| = |q₃| · E = 40·10⁻⁶ · 150 = 6·10⁻³ N
m =
|F|
a
=
6·10⁻³
2
= 3·10⁻³
kg
3
▾
✕
I (A) : 0.5 1.0 1.5 2.0 2.5
V (V) : 4.9 3.9 3.2 2.0 0.8
40 %
4.00
1.00
4.00
6.00
2.00
I =
ε
r + R
MK
(0.5 , 4.9) (1.0 , 3.9) (1.5 , 3.2) (2.0 , 2.0) (2.5 , 0.8)
V = −2.02 · I + 5.99
V = ε − I·r
V = ε − I·r
I = 0 ⟹ V = ε
ε = 6.0
V
ε = V + I·r = 5.0 + 0.5 · 2.0 = 6.0
V
(I₁ , V₁) = (0.5 , 5.0) , (I₂ , V₂) = (2.5 , 1.0)
=
V₂ − V₁
I₂ − I₁
=
1.0 − 5.0
2.5 − 0.5
=
−4.0
2.0
= −2.0
Ω
r = 2.0
Ω
R
MK
= 0
I =
ε
r + R
MK
=
6.0
2.0 + 0
= 3.0
A
R =
V
I
=
R = 9.80 3.90 2.13 1.00 0.32
Ω
V = ε − I·r
4
▾
✕
I = 8.5 A | r = 10 cm | α = 37° | μ₀ = 4π·10⁻⁷ T·m/A
8.5 A
1.70
2.26
37
0.75
B
I
=
μ₀·I
2π·r
tan(α) =
B
I
B
E
tan(α) =
μ₀·I
2π·r·B
E
B
I
=
4π·10⁻⁷ · 8.5
2π · 0.10
=
2·10⁻⁷ · 8.5
0.10
= 1.7·10⁻⁵
T
B
E
=
B
I
tan(37°)
=
1.7·10⁻⁵
0.754
= 2.26·10⁻⁵
T
tan(55°) = 1.43 > 1 ⟹ B
I
> B
E
5
▾
✕
B = 1 T | v = 3.6·10⁶ m/s | m = 6.67·10⁻²⁷ kg
PC = 15 cm | QF = 15 cm | e = 1.6·10⁻¹⁹ C
15.0
1.60
0.0
0.00
0
F = q · v × B
: •
: ×
: ×
Δx = 15 cm , Δy = 15 cm ⟹ r = 0.15 m
r =
m·v
|q|·B
⟹ |q| =
m·v
r·B
|q| =
6.67·10⁻²⁷ · 3.6·10⁶
0.15 · 1
=
2.40·10⁻²⁰
0.15
= 1.6·10⁻¹⁹ C
W = 0 ⟹ ΔE
k
= 0 ⟹ v =
90° + 180° + 90° = 360°
S = 2π·r = 2π · 0.15 = 0.942
m
t =
S
v
=
0.942
3.6·10⁶
= 2.6·10⁻⁷
s
T =
2π·m
|q|·B
=
2π · 6.67·10⁻²⁷
1.6·10⁻¹⁹ · 1
= 2.6·10⁻⁷
s
F = q · v × B
6
▾
✕
K = 0.02 | R = 0.04 Ω | a = 2 m/s²
B(x) = B
0,x
− K·x | x = ½·a·t²
2.0 m/s²
0.020 T/m
0.00
0.0
5.00
10.0
0.00
B(x) = B
0,x
− K·x
K =
B
0,x
− B(x)
x
⟹ [K] =
T
m
Φ = B · A
ε = −
ΔΦ
Δt
, I =
ε
R
x(t) =
1
2
·a·t²
B(t) = B
0,x
− K ·
1
2
·a·t²
Φ(t) = A · B(t) = A·B
0,x
−
1
2
·A·K·a·t²
|ε| =
ΔΦ
Δt
= A·K·a·t
I(t) =
|ε|
R
=
A·K·a
R
· t
=
0.6 − 0
0.3 − 0
= 2.0
mA/s
= 2·10⁻³
A/s
A·K·a
R
= 2·10⁻³ ⟹ A =
2·10⁻³ · R
K · a
A =
2·10⁻³ · 0.04
0.02 · 2
=
8·10⁻⁵
0.04
= 2·10⁻³
m²
→
🏠
♿