Download Problem 4.31 The circular disk of radius a shown in Fig. 4

EE324
HW7
Spring12
Problem 4.31 The circular disk of radius a shown in Fig. 4-7 has uniform charge
density ρs across its surface.
(a) Obtain an expression for the electric potential V at a point P = (0, 0, z) on the
z-axis.
(b) Use your result to find E and then evaluate it for z = h. Compare your final
expression with (4.24), which was obtained on the basis of Coulomb’s law.
Solution:
z
E
P(0,0,h)
h
ρs
dq = 2π ρs r dr
r
a
y
dr
a
x
Figure P4.31: Circular disk of charge.
(a) Consider a ring of charge at a radial distance r. The charge contained in
width dr is
dq = ρs (2π r dr) = 2πρs r dr.
The potential at P is
dV =
2πρs r dr
dq
.
=
4πε0 R 4πε0 (r2 + z2 )1/2
The potential due to the entire disk is
V=
Z a
0
dV =
ρs
2ε0
Z a
0
¯
ρs 2 2 1/2 ¯¯a
ρs h 2 2 1/2 i
r dr
=
=
(r
+
z
)
(a + z ) − z .
¯
2ε0
2ε0
(r2 + z2 )1/2
0
(b)
·
¸
∂V
∂V
∂V
ρs
z
E = −∇V = −x̂
− ŷ
− ẑ
= ẑ
1− √
.
∂x
∂y
∂z
2ε0
a2 + z2
The expression for E reduces to Eq. (4.24) when z = h.
Problem 4.34 Find the electric potential V at a location a distance b from the origin
in the x–y plane due to a line charge with charge density ρℓ and of length l. The line
charge is coincident with the z-axis and extends from z = −l/2 to z = l/2.
Solution: From Eq. (4.48c), we can find the voltage at a distance b away from a line
z
|R'| = z2 + b2
l/2
dz
z
l
R'
V(b)
b
y
-l/2
Figure P4.34: Line of charge of length ℓ.
of charge [Fig. P4.34]:
1
V (b) =
4πε
Z
l′
ρl ′
ρl
dl =
′
R
4πε
dz
ρl
√
=
ln
2
2
4πε
−l/2
z +b
Z l/2
Ã
!
√
l + l 2 + 4b2
√
.
−l + l 2 + 4b2
Problem 4.38 Given the electric field
E = R̂
18
R2
(V/m)
find the electric potential of point A with respect to point B where A is at +2 m and
B at −4 m, both on the z-axis.
Solution:
A
z = 2m
B
z = -4m
Figure P4.38: Potential between B and A.
VAB = VA −VB = −
Along z-direction, R̂ = ẑ and E = ẑ
z ≤ 0. Hence,
VAB = −
Z 2
−4
R̂
18
· ẑ dz = −
z2
Z A
B
E · dl.
18
18
for z ≥ 0, and R̂ = −ẑ and E = −ẑ 2 for
2
z
z
·Z
0
−4
−ẑ
18
· ẑ dz +
z2
Z 2
18
ẑ
0
z2
¸
· ẑ dz = 4 V.
Problem 4.48 With reference to Fig. 4-19, find E1 if E2 = x̂3 − ŷ2 + ẑ2 (V/m),
ε1 = 2ε0 ,
ε2 = 18ε0 , and the boundary has a surface charge density
ρs = 3.54 × 10−11 (C/m2 ). What angle does E2 make with the z-axis?
Solution: We know that E1t = E2t for any 2 media. Hence, E1t = E2t = x̂3 − ŷ2.
Also, (D1 − D2 ) · n̂ = ρs (from Table 4.3). Hence, ε1 (E1 · n̂) − ε2 (E2 · n̂) = ρs , which
gives
E1z =
ρs + ε2 E2z 3.54 × 10−11 18(2)
3.54 × 10−11
=
+
=
+ 18 = 20 (V/m).
ε1
2ε0
2
2 × 8.85 × 10−12
Hence, E1 = x̂3 − ŷ2 + ẑ20 (V/m). Finding the angle E2 makes with the z-axis:
¶
µ
√
2
−1
√
= 61◦ .
2 = 9 + 4 + 4 cos θ ,
θ = cos
E2 · ẑ = |E2 | cos θ ,
17
Problem 4.55 In a dielectric medium with εr = 4, the electric field is given by
E = x̂(x2 + 2z) + ŷx2 − ẑ(y + z) (V/m)
Calculate the electrostatic energy stored in the region −1 m ≤ x ≤ 1 m, 0 ≤ y ≤ 2 m,
and 0 ≤ z ≤ 3 m.
Solution: Electrostatic potential energy is given by Eq. (4.124),
1
ε 3 2
ε |E| d V =
[(x2 + 2z)2 + x4 + (y + z)2 ] dx dy dz
2 z=0 y=0 x=−1
V
¯ ¯

µ
¶ ¯¯1
¯3
¯2
4ε0  2 5
2 2 3
4 3
1
¯ ¯
¯
4

=
x yz + z x y + z xy + (y + z) x ¯
¯
¯
¯
¯
¯
2
5
3
3
12
z=0
y=0
x=−1
µ
¶
4ε0 1304
=
= 4.62 × 10−9 (J).
2
5
1
We =
2
Z
2
Z
Z
Z
Problem 4.58 The capacitor shown in Fig. P4.58 consists of two parallel dielectric
layers. Use energy considerations to show that the equivalent capacitance of the
overall capacitor, C, is equal to the series combination of the capacitances of the
individual layers, C1 and C2 , namely
C=
where
C1 = ε1
C1C2
C1 +C2
A
,
d1
C2 = ε2
(22)
A
d2
(a) Let V1 and V2 be the electric potentials across the upper and lower dielectrics,
respectively. What are the corresponding electric fields E1 and E2 ? By
applying the appropriate boundary condition at the interface between the two
dielectrics, obtain explicit expressions for E1 and E2 in terms of ε1 , ε2 , V , and
the indicated dimensions of the capacitor.
(b) Calculate the energy stored in each of the dielectric layers and then use the sum
to obtain an expression for C.
(c) Show that C is given by Eq. (22).
d1
A
d2
+
−
ε1
V
ε2
(a)
C1
+
V
−
C2
(b)
Figure P4.58: (a) Capacitor with parallel dielectric
layers, and (b) equivalent circuit (Problem 4.58).
Solution:
+
d1
ε1
d2
ε1
E1
+
V1
-
+
E2 V1
-
V
Figure P4.58: (c) Electric fields inside of capacitor.
(a) If V1 is the voltage across the top layer and V2 across the bottom layer, then
V = V1 +V2 ,
and
E1 =
V1
,
d1
E2 =
V2
.
d2
According to boundary conditions, the normal component of D is continuous across
the boundary (in the absence of surface charge). This means that at the interface
between the two dielectric layers,
D1n = D2n
or
ε1 E1 = ε2 E2 .
Hence,
V = E1 d1 + E2 d2 = E1 d1 +
which can be solved for E1 :
E1 =
V
ε1 .
d1 + d2
ε2
Similarly,
E2 =
V
ε2 .
d2 + d1
ε1
ε1 E1
d2 ,
ε2
(b)
2

·
¸
1 2
1
ε1 ε22 Ad1
1 
V

2
We1 = ε1 E1 · V 1 = ε1 
ε1  · Ad1 = 2 V (ε2 d1 + ε1 d2 )2 ,
2
2
d1 + d2
ε2
2

¸
·
1 2
1
ε12 ε2 Ad2
1 
V

2
We2 = ε2 E2 · V 2 = ε2 
ε2  · Ad2 = 2 V (ε1 d2 + ε2 d1 )2 ,
2
2
d2 + d1
ε
· 2 1 2
¸
1 2 ε1 ε2 Ad1 + ε1 ε2 Ad2
We = We1 +We2 = V
.
2
(ε1 d2 + ε2 d1 )2
But We = 12 CV 2 , hence,
C=
ε1 ε22 Ad1 + ε12 ε2 Ad2
ε1 ε2 A
(ε2 d1 + ε1 d2 )
= ε1 ε2 A
=
.
2
2
(ε2 d1 + ε1 d2 )
(ε2 d1 + ε1 d2 )
ε2 d1 + ε1 d2
(c) Multiplying numerator and denominator of the expression for C by A/d1 d2 , we
have
ε1 A ε2 A
·
C1C2
d
d2
=
,
C= 1
ε1 A ε2 A C1 +C2
+
d1
d2
where
C1 =
ε1 A
,
d1
C2 =
ε2 A
.
d2
Problem 4.62 Conducting wires above a conducting plane carry currents I1 and I2
in the directions shown in Fig. P4.62. Keeping in mind that the direction of a current
is defined in terms of the movement of positive charges, what are the directions of
the image currents corresponding to I1 and I2 ?
I2
I1
(a)
(b)
Figure P4.62: Currents above a conducting plane
(Problem 4.62).
Solution:
(a) In the image current, movement of negative charges downward = movement of
positive charges upward. Hence, image of I1 is same as I1 .
+ q @ t=t1
I1
+ q @ t=0
I1
(image)
- q @ t=0
- q @ t=t1
Figure P4.62(a): Solution for part (a).
(b) In the image current, movement of negative charges to right = movement of
positive charges to left.
I1
@t=0
+q
+ q @ t=t1
-q
@t=0
- q @ t=t1
I1
(image)
Figure P4.62(b): Solution for part (b).
z
1m
5A
4 cm
y
-1 m
Figure P5.5: Problem 5.5.
Problem 5.5 In a cylindrical coordinate system, a 2-m-long straight wire carrying
a current of 5 A in the positive z-direction is located at r = 4 cm, φ = π /2, and
−1 m ≤ z ≤ 1 m.
(a) If B = r̂ 0.2 cos φ (T), what is the magnetic force acting on the wire?
(b) How much work is required to rotate the wire once about the z-axis in the
negative φ -direction (while maintaining r = 4 cm)?
(c) At what angle φ is the force a maximum?
Solution:
(a)
F = Iℓℓ × B
× [r̂ 0.2 cos φ ]
= 5ẑ 2×
= φ̂φ 2 cos φ .
At φ = π /2, φ̂φ = −x̂. Hence,
F = −x̂ 2 cos(π /2) = 0.
(b)
Z 2π
Z 2π
¯
¯
W=
F · dl =
φ̂φ [2 cos φ ] ·(−φ̂φ)r d φ ¯¯
0
φ =0
r=4 cm
¯
Z 2π
¯
= −2r
= −8 × 10−2 [sin φ ]20π = 0.
cos φ d φ ¯¯
0
r=4 cm
The force is in the +φ̂φ-direction, which means that rotating it in the −φ̂φ-direction
would require work. However, the force varies as cos φ , which means it is positive
when −π /2 ≤ φ ≤ π /2 and negative over the second half of the circle. Thus, work
is provided by the force between φ = π /2 and φ = −π /2 (when rotated in the
−φ̂φ-direction), and work is supplied for the second half of the rotation, resulting in a
net work of zero.
(c) The force F is maximum when cos φ = 1, or φ = 0.
Problem 5.12 Two infinitely long, parallel wires are carrying 6-A currents in
opposite directions. Determine the magnetic flux density at point P in Fig. P5.12.
I1 = 6 A
I2 = 6 A
P
0.5 m
2m
Figure P5.12: Arrangement for Problem 5.12.
Solution:
B = φ̂φ
µ0 I1
µ0 I2
µ0
8µ0
(6 + 2) = φ̂φ
+ φ̂φ
= φ̂φ
2π (0.5)
2π (1.5)
π
π
(T).
Problem 5.19 Three long, parallel wires are arranged as shown in Fig. P5.19.
Determine the force per unit length acting on the wire carrying I3 .
I1 = 10 A
2m
I3 = 10 A
2m
2m
I2 = 10 A
Figure P5.19: Three parallel wires of Problem 5.19.
Solution: Since I1 and I2 are equal in magnitude and opposite in direction, and
z
I1 x
into the page (y)
R1
2m
B2
R2
x
I3
B1
2m
x
R1
R2 = R1
I2
out of the page (-y)
Figure P5.19: (a) B fields due to I1 and I2 at location of I3 .
I1 x
F'31
'
F32
θ
x
x
I3
I2
Figure P5.19: (b) Forces acting on I3 .
equidistant from I3 , our intuitive answer might be that the net force on I3 is zero.
As we will see, that’s not the correct answer. The field due to I1 (which is along ŷ) at
location of I3 is
µ0 I1
B1 = b̂1
2π R1
where b̂1 is the unit vector in the direction of B1 shown in the figure, which is
perpendicular to R̂1 . The force per unit length exerted on I3 is
F′31 =
µ0 I1 I3
µ II
× b̂1 ) = −R̂1 0 1 3 .
(ŷ×
2π R1
2π R1
Similarly, the force per unit length excited on I3 by the field due to I2 (which is
along −ŷ) is
µ0 I2 I3
F′32 = R̂2
.
2π R2
The two forces have opposite
components along
components
along ẑ.
√
√ x̂ and equal
√
−1
−1
◦
Hence, with R1 = R2 = 8 m and θ = sin (2/ 8) = sin (1/ 2) = 45 ,
¶
µ
µ0 I1 I3 µ0 I2 I3
′
′
′
+
sin θ
F3 = F31 + F32 = ẑ
2π R1
2π R2
¶
µ
4π × 10−7 × 10 × 20
1
√
= ẑ 2
× √ = ẑ 2 × 10−5 N/m.
2π × 8
2
Problem 5.20 A square loop placed as shown in Fig. P5.20 has 2-m sides and
carries a current I1 = 5 A. If a straight, long conductor carrying a current I2 = 10 A is
introduced and placed just above the midpoints of two of the loop’s sides, determine
the net force acting on the loop.
z
I2
4
1
a
a
3
y
I1
2
x
Figure P5.20: Long wire carrying current I2 , just above
a square loop carrying I1 (Problem 5.20).
Solution: Since I2 is just barely above the loop, we can treat it as if it’s in the same
plane as the loop. For side 1, I1 and I2 are in the same direction, hence the force on
side 1 is attractive. That is,
F1 = ŷ
µ0 I1 I2 a
4π × 10−7 × 5 × 10 × 2
= ŷ
= ŷ 2 × 10−5 N.
2π (a/2)
2π × 1
I1 and I2 are in opposite directions for side 3. Hence, the force on side 3 is repulsive,
which means it is also along ŷ. That is, F3 = F1 .
The net forces on sides 2 and 4 are zero. Total net force on the loop is
F = 2F1 = ŷ 4 × 10−5 N.