HW # 2: 4.29, 4.36, 4.44, 4.54, 4.60, 4.63 - Due Sept. 23 ... Solutions.pdf · HW # 2: 4.29, 4.36, 4.44, 4.54, 4.60, 4.63 - Due Sept. 23, 2011 Solutions 4.29 A spherical shell with
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HW # 2: 4.29, 4.36, 4.44, 4.54, 4.60, 4.63 - Due Sept. 23, 2011
Solutions
4.29 A spherical shell with outer radius b surrounds a charge-free cavity of radius a < b (Fig. P4.29).
If the shell contains a change density given by
, where is a positive
constant, determine D in all regions.
This should be broken up into three regions. Region 1 includes Region 2 includes
Region 3 includes For each region, ∮
. Since D is radially oriented,
| |, so ∮
∫ | |
| | ∫
| | ∫
| |
For region 1,
For region 2, ∫
, where r’ is the location of interest.
For region 3, ∫
Therefore, for any r, {
( )
( ; )
4.36 For each of the distributions of the electric potential V shown in Fig P4.36, sketch the
corresponding distribution of E (in all cases, the vertical axis is in volts and the horizontal axis is in
meters).
Given 𝐸 ∇𝑉, basically each plot will be the piecewise slope of the given plot:
4.44 A coaxial resistor of length l consists of two concentric cylinders. The inner cylinder has radius
a and is made of a material with conductivity , and the outer cylinder extending between r = a
and r = b, is made of a material with conductivity . If the two ends of the resistor are capped
with conducting plates, show that the resistance between the two ends is
[ ( (
))]⁄ .
Because the plate covers the entire surface of the ends of each of these cylinders, one needs only
find the resistance of each cylinder independently, and add the resistances in parallel. Given
𝑅 𝑙
𝜎𝐴 and 𝐴 𝜋𝑟 , 𝑅𝑖𝑛𝑛𝑒𝑟
𝑙
𝜎1𝜋𝑎 and 𝑅𝑜𝑢𝑡𝑒𝑟
𝑙
𝜎 (𝜋𝑏 ;𝜋𝑎 ). The total resistance, then:
1
𝑅 𝑅𝑖𝑛𝑛𝑒𝑟 𝑅𝑜𝑢𝑡𝑒𝑟𝑅𝑖𝑛𝑛𝑒𝑟 𝑅𝑜𝑢𝑡𝑒𝑟
𝑙𝜍 𝜋𝑎
𝑙
𝜍 (𝜋𝑏 𝜋𝑎 )
𝑙𝜍 𝜋𝑎
𝑙𝜍 (𝜋𝑏 𝜋𝑎 )
𝑙𝜍 (𝜋𝑏 𝜋𝑎 ) 𝑙𝜍 𝜋𝑎
𝜍 𝜋𝑎 𝜍 (𝜋𝑏
𝜋𝑎 )
𝑙
𝜍 𝜋𝑎 𝜍 (𝜋𝑏
𝜋𝑎 )
𝑅 𝑙
𝑙𝜍 (𝜋𝑏 𝜋𝑎 ) 𝑙𝜍 𝜋𝑎
𝑙
𝜋(𝜍 𝑏 𝜍 𝑎
𝜍 𝑎 )
𝑙
𝜋(𝜍 𝑎 𝜍 (𝑏
𝑎 )
4.54 An electron with charge qc = -1.6x10-19 C and mass me=9.1x10-31kg is injected at a point
adjacent to the negatively charged plate in the region between the plates of an air-filled parallel
plate capacitor with separation of 1cm and rectangular plates each 10cm2 in area. If the voltage
across the capacitor is 10V, find the following:
a) The force acting on the electron.
b) The acceleration of the electron.
c) The time it takes the electron to reach the positively charged plate, assuming that it starts from
rest.
a) 𝐅 𝑞E, and from eq 4.112, we know that the Voltage between two parallel plates is
𝑉 𝐸𝑑 so F 𝑞𝑉
𝑑
6 −19
1 6 1 ; 6
b) F 𝑚a, so 𝑎 6 −16
9 31 1 76 1 4
c) 𝑑 𝑎𝑡
, so 𝑡
𝑑
𝑎
76 14 1 7 1 8
4.60 A coaxial capacitor consists of two concentric conducting cylindrical surfaces, one of a radius
a and another of radius b, as shown in Fig P4.60. The insulating layer separating the two
conducting surfaces is divided equally into two semi cylindrical sections, one filled with dielectric
ε1 and the other filled with dielectric ε2.
a) Develop an expression for C in terms of the length l and the given quantities.
b) Evaluate the value of C of a = 2mm, b = 6mm, εr1 = 2, εr2 = 4, and l = 4cm.
a) Let E1 be the field inside the dielectric ε1 and E2 be the field in dielectric ε2. At the interface
of the two, boundary conditions require that E1 = E2. At 𝑟 𝑎 (surface of inner conductor,
in medium 1, the boundary condition on D, as stated by eq 4.101, leads to
𝜀 E n𝜌𝑠 and 𝜀 𝐸 𝜌𝑠 , or 𝜌𝑠 𝜀 𝐸. Similarly, in medium 2, 𝜌𝑠 𝜀 𝐸.
Thus, the E fields will be the same in the two dielectrics, but the charge densities will be
different along the two sides of the inner conducting cylinder.
Since the same voltage applies for the two sections of the capacitor, we can treat them
as two capacitors in parallel. For each capacitor, we apply the results of Equations 4.114
through 4.116, except for only one-half of a full cylinder each: 𝐶𝑥 𝜋𝜀𝑥𝑙
ln(𝑏 𝑎 ), and
𝐶 𝐶 𝐶 𝜋𝑙(𝜀 𝜀 )
ln(𝑏 𝑎 )
b) 𝐶 𝜋 4 − ( :4) 8 85 −1
ln(6 ) 6 7𝑝𝐹
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