南京大学大学物理答案15-16

0
(
)
(c) In (a) VP(0, y) . For Ex we need V(x, y)
Problem 16.1 before after
A
C=
ε0 A
d
b
copper
d
ε0 A C1 = C2 = ( d − b) 2
d −b
in series C = 1 C = ε 0 A 1
2
Problem 16.3
(a)
1 ⌠ λ dx VP = 2 4πε 0 ⌡ x + y 2
0
L
(
)
12
k 2 2 = L + y − y 4πε 0
(b)
dV k Ey = − = 1− dy 4πε 0
L
y 2 2 L +y
1 ⌠ yλ dx Ey = =⋯ 2 32 4πε 0 ⌡ x + y 2
q −αx E= = e ε 0ε r A ε 0 A
⌠ q −αx − V = − e dx ⌡0 ε 0 A
d
q
Vα ε 0 A q= −α d 1− e
E=
αV
1− e
−α d
e
−αx
Problem 16.6 Gauss’ law reads 16.4
2 π ρℓ ⋅ E ρ = λℓ ε 0
Problem 16.9 In the case q = const. double separation means half the capacitance and double energy. (a)
0.20 × 10−4 C= = 9 d 4π × 9 × 10 × 0.010
ε0 A
= 2.0 × 10
Problem 15.1 Equilibrium requires that all forces are cancelled.
F on q
− αq 4q + =0 2 2 d (d − x )
2 2 2 2
4αq αq F on − αq − 2 + =0 2 x (d − x )
F on 4q 4αq − 4q + =0 2 2 x d
Problem 15.3 Directly use the result of dipole NB p = qa
2p 1 1 EP = − 3 3 4πε 0 a a r− r+ 2 2
P
r
q a –2q a q
2 p 3a ⋅ r 3 2qa 2 ˆ ≈ ⋅ 5 = ⋅ 4 a 4πε 0 r 4πε 0 r
−12
× 100C = 8.8 × 10
−10
C
(c)In the gap
C0 ∆V q 4 V Egap = = = 1.0 × 10 ε 0 A C0 d m
(d)
∆V = E gap ⋅ 0.50cm + Ed ⋅ 0.50cm = 57V
(e)
u gap : u d =
ε0
2
2 Egap
ε0 2 ε0 1 2 1 Egap : Ed ⋅ D = Egap : 2 2 εr 2
Problem 15.6 Define R = cos ϕ i + sin ϕ j
1 ⌠ λ ⋅ Rdϕ (− R ) E= 2 4πε 0 ⌡ R
1 ⌠ − λ ⋅ Rdϕ (− R ) + 2 4πε 0 ⌡ R
π2
π2
0 π
1 4Q (− i ) = 2 4πε 0 πR
Problem 15.9 An infinite sheet of dx' at x' contributes a field
The potential difference is
R′
λ Eρ = 2πε 0 ρ
λ R′ V ( R ) − V ( R ′) = E ρ dρ = ln 2 πε 0 R
∫
R
The capacity in unit length is
outside wall is equivalent a little larger R', so C will be smaller a little bit.
2 2
2d x= ; α =4 9 3 Assume one of these charges, say q, moves a small ∆
4q αq Fq = − ~ +∆ 2 2 (d + ∆) (d − x + ∆)
2 2
not restoring force. Unstable. Earshaw’s Theorem: A charge particle cannot be held in a stable equilibrium by electrostatic forces alone.
at both sides. So the field due to other part than dA is 1 1 2 πσ = E outward.
4 πε 0
2
dA
total ?
Other part
The force on dq is then
1 dF = E dq 2
Problem 15.15
or 87.5% and 12.5%
1 = 1: 7
2
−14
,
−11 2
q = 2.0 × 10
−11
C
q 2.0 × 10 U= = 2C 2 × 2.0 × 10−14
(
)
= 1.0 × 10 J
−8
(b)
1 1 1 C0 2 2 2 (∆V ) C0 (1000) = C1 (∆V ) = 2 2 2 2
∆V = 2000V
Problem 16.4 A dielectric slab of thickness b 16.10 and dielectric constant is placed between the plates of a parallel-plate capacitor. Known
Outside the sheet, E = 0
*Problem 15.12 Total field is
1 Q 1 E= = 4 πσ 4 πε 0 r 2 4 πε 0
The contribution from dA (very close to the surface) is
1 2 πσ 4 πε 0
1 E= 2π(− eax′)dx′ 4πε 0
left to x
1 1 E ( x) = 2π (−eax′)dx′ − 2π 4πε 0 − x 2 4πε 0
∫
m
x
xm 2
∫
x
(−eax′)dx′
2 x 2 x 2 x 2 ea x m m =− − + − − 2ε 0 2 4 4 2 x 2 ea m − x2 = 2ε 0 2
A = 100cm , d = 1.0cm,
2
ε r = 7.0, V0 = 100V, b = 0.50cm
(a)
100 × 10 −4 C0 = = F = 8.8 × 10 −12 F 9 −2 d 4π × 9 × 10 × 1.0 × 10 (b)
ε0 A
q = C 0 ⋅ ∆V = 8.8 × 10
a
⌠ dx ε 0a ⌠ ε 0 adx ≈ C= d 1+ θ x ⌡ d + x tanθ ⌡ 0 d a 0 ε 0a ⌠ θ = 1 − d x + ⋯ dx d ⌡
0
a
aθ = 1 − d 2d
2
ε 0a
θ
d a
Problem 16.6 Known V