Untitled OmniPage Document - University of Arizona

Problem 4.48 With reference to Fig. 4-19, find E1 if E2 = x^3 - y^2 + z^2 (V/m), 1 = 20, 2 = 180, and the boundary has a surface charge density s = 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 = x^3 - y^2. Also, (D1 - D2) ? n^ = s (from Table 4.3). Hence, 1(E1 ? n^ ) - 2(E2 ? n^ ) = s, which gives

E1z

=

s

+ 2E2z 1

=

3.54 ? 10-11 20

+

18(2) 2

=

3.54 ? 10-11 2 ? 8.85 ? 10-12

+ 18

=

20

(V/m).

Hence, E1 = x^3 - y^2 + z^20 (V/m). Finding the angle E2 makes with the z-axis:

E2 ? z^ = |E2| cos ,

2 = 9 + 4 + 4 cos ,

= cos-1 2 = 61. 17

Problem 4.56 Figure P4.56(a) depicts a capacitor consisting of two parallel, conducting plates separated by a distance d. The space between the plates contains two adjacent dielectrics, one with permittivity 1 and surface area A1 and another with 2 and A2. The objective of this problem is to show that the capacitance C of the configuration shown in Fig. P4.56(a) is equivalent to two capacitances in parallel, as illustrated in Fig. P4.56(b), with

C = C1 +C2

(19)

where

C1

=

1A1 d

(20)

C2

=

2A2 d

(21)

To this end, proceed as follows:

(a) Find the electric fields E1 and E2 in the two dielectric layers.

(b) Calculate the energy stored in each section and use the result to calculate C1 and C2.

(c) Use the total energy stored in the capacitor to obtain an expression for C. Show that (19) is indeed a valid result.

A1

A2

+

V

-

d

1

2

(a)

+

C1

C2

V -

(b)

Figure P4.56: (a) Capacitor with parallel dielectric section, and (b) equivalent circuit.

Solution:

1

2

E1 E2

d

+ -V

(c)

Figure P4.56: (c) Electric field inside of capacitor.

(a) Within each dielectric section, E will point from the plate with positive voltage to the plate with negative voltage, as shown in Fig. P4-56(c). From V = Ed,

E1

=

E2

=

V d

.

(b)

We1

=

1 2

1E12

?

V

=

1 2

1

V d

2 2

? A1d

=

1 2

1V

2

A1 d

.

But, from Eq. (4.121),

We1

=

1 2

C1V 2.

Hence

C1

=

1

A1 d

.

Similarly,

C2

=

2

A2 d

.

(c) Total energy is

Hence,

We

= We1

+ We2

=

1 2

V2 d

(1A1

+ 2A2)

=

1 CV 2. 2

C

=

1A1 d

+

2A2 d

= C1 +C2.

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

P 0.5 m

I2 = 6 A

2 m

Figure P5.12: Arrangement for Problem 5.12.

Solution:

B = ^

?0I1 2 (0.5)

+

^

?0I2 2 (1.5)

= ^

?0

(6

+

2)

=

^

8?0

(T).

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