PHYS3041 PHYS3041 ELECTRICITY ELECTRICITY AND MAGNETISM MAGNETISM HOMEWOR HOMEWORK K 2 SOLUTI SOLUTION ON
M.T.C. WONG
Abstract. Questions 1 and 2 are about work and electrostatic electrostatic energy. Various forms of electrostatic energy density can be used during integration. Question 3 and 4 are about conductors. conductors. Different Different geometry geometry of physical physical system are specified in problems, which lead to different forms of capacitance and treatment of boundary conditions.
1. Problem 2.32 of Griffiths Find the energy stored in a uniformly charged solid sphere of radius R and charge q. (a) You found the potential in Problem 2.21: 1 q V in in = 4π0 2R
(1.1)
3−
r2 R2
Use equation 2.43: (1.2) (1.3) (1.4) (1.5) (1.6)
1 W = 2 1 = 2
ρV dτ
R
0
1 q ρ 4π0 2R
3q 2 = 16 16π π0 R4 3q 2 = 20 20π π0 R 1 3q 2 = 4π 0 5R
R
3r
0
3−
2
−
r2 R2 r4 R2
4πr 2 dr
dr
(b) Find the E-field inside and outside the sphere:
out E out
(1.8)
Date :
1 q r rˆ 4π0 R3 1 q = rˆ 4π0 r2
in E in =
(1.7)
September 27, 2010. 1
2
M.T.C. WONG
Use equation 2.45: 0 W = 2
(1.9)
E 2 dτ
all space
0 = 2
(1.10)
R
2 E in
0
R
2
q = 8π0
(1.11)
0
∞
0 dτ + 2
2 E out dτ
R
∞
4
r dr + R6
R
1 dr r2
2
q 1 1 + 8π0 5R R 1 3q2 = 4π0 5R
(1.12)
=
(1.13)
(c) Find the potential outside the sphere: (1.14)
1 q 4π 0 r
V out =
Use equation 2.44: (1.15)
W =
0 2
E 2 dτ +
V
(1.16) (1.17) (1.18)
· da V E
S
R
0 0 2 = E in dτ + 2 0 2 2 q 6 1 1 = + − 8π0 5R a a 1 3q2 = 4π0 5R
a
When a → ∞, we will have
0 2
2 E out
R
0 dτ + 2
E 2 dτ →
V
0 2
1 q 4π 0 a
E 2 dτ , and
all space
2. Problem 2.33 of Griffiths When the sphere has radius r, the charge is q . r3 R3 3q (2.2) dq = 3 r2 dr R Amount of work dW taken to build up the radius: q = q
(2.1)
(2.3) (2.4)
1 q dq 4π0 r 1 q r3 = 4π0 r R3
dW =
Integrate this to find the work:
3qr 2 R3
1 q 4π0 a2
dr
0 2
4πa2
· da → 0. V E
S
PH YS 30 41 EL EC TR IC IT Y A ND M AGN ET IS M
(2.5)
W =
(2.6)
=
H OME WO RK 2 S OL UT IO N
dW
R
(2.7)
3q 2 r4 dr 4π 0 R6 0 1 3q 2 = 4π0 5R 3. Problem 2.39 of Griffiths
Let the charge be Q when considering the length of the tube being L. (3.1) (3.2)
· da = Q E 0 Q E (2πrL) = 0
(3.3)
E =
Q 2π0 Lr
The potential difference is then: b
(3.4)
∆V =
· d E l
a b
Q dr a 2π 0 Lr Q b (3.6) = ln 2π 0 L a So the capacitance per unit length: (3.5)
(3.7) (3.8)
=
C/L =
Q ∆V
L 2π0 = ln b/a
4. Problem 2.40 of Griffiths (a) Use equation 2.52, which is about electrostatic pressure: (4.1)
P =
0 2 E 2
The work done by electrostatic forces: (4.2)
W = force × distance
(4.3)
= (P × A) × 0 = E 2 A 2
(4.4)
3
4
M.T.C. WONG
(b) Use equation 2.46, the energy per unit volume is by the field:
0 2 E . 2
So the energy lost
(4.5)
W = energy density × volume 0 2 (4.6) = E A 2 Therefore the energy lost by the field is the same as the work done by electrostatic forces in part (a). The following section gives an alternative derivation of equation 2.51 of Griffiths: The surface has a finite thickness d, and the charge density is a constant ρ. Within the thickness of the patch, let the area of the patch be A, and the distance away from the inner surface of the conductor be r. There is no electric field found inside the conductor, i.e. E = 0 at r = 0. The electric field at other values of r is: · da = ρ × volume (4.7) E 0 ρAr (4.8) E · A = 0 ρr (4.9) E = 0 By considering the integral, the total force acting on the patch is:
(4.10)
Total force =
E dq
d
(4.11)
=
0
ρr 0
(ρAdr)
d
ρ2 A (4.12) = r dr 0 0 ρ2 d2 A (4.13) = 20 When d → 0 and ρ → ∞ such that ρd → σ, the total force becomes the following form as in equation 2.51: Force ρ2 d2 = lim ρd→σ 20 Area σ2 (4.15) = 20 And finally the pressure on the patch is: (4.14)
(4.16) (4.17)
σ2 20 0 = E 2 2
P =
Department of Physics, The Chinese University of Hong Kong, Hong Kong, China E-mail address :
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