Problem 4.31 The circular disk of radius Obtain an expression for the ...
EE324
HW7
Spring12
Problem 4.31 The circular disk of radius a shown in Fig. 4-7 has uniform charge density s 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
s
dq = 2 s r dr
r
ay
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 = s(2r dr) = 2sr dr.
The potential at P is
dV
=
dq 4 0 R
=
2sr dr 40(r2 + z2)1/2
.
The potential due to the entire disk is
V=
a
dV
0
=
s 20
a 0
r dr (r2 + z2)1/2
=
s 20
(r2 + z2)1/2
a
=
0
s 20
(a2 + z2)1/2 - z .
(b)
E = -V
=
-x^
V x
-
y^
V y
-
z^
V z
=
z^
s 20
1- z 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 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
l/2
dz
l
z
|R'| = z2 + b2
R'
V(b)
y
b
-l/2
Figure P4.34: Line of charge of length .
of charge [Fig. P4.34]:
V
(b)
=
1 4
l
l R
dl
=
l 4
l/2 -l/2
dz z2 + b2
=
l 4
ln
l + l2 + 4b2 -l + l2 + 4b2
.
Problem 4.38 Given the electric field
E
=
R^
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.
A
VAB = VA -VB = - E ? dl.
B
Along
z-direction,
R^
=
z^
and
E
=
z^
18 z2
for
z
0,
and
R^
=
-z^
and
E
=
-z^
18 z2
for
z 0. Hence,
VAB = -
2 -4
R^
18 z2
?
z^
d
z
=
-
0 -4
-z^
18 z2
?
z^
d
z
+
2 0
z^
18 z2
?
z^
dz
= 4 V.
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
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