2, Homework Physics 604
Prof. Charles E. Hyde-Wright
(Dated: Due: Sept 26, 2006)
2.1. A grounded, metal sphere of radius R is centered at the origin. The sphere is in the presence of an electric eld with asymptotic boundary condition E(x, y, z) E0 z as r . (1)
(a) Show that the electrostatic potential for r > R can be obtained by superposing the dipole eld Ext (x, y, z) = E0 z (2)
with the potential from an image-dipole at the origin. The image-dipole is a pair of charges q located at z = a. The solution is obtained in the following limit: a0 q qa = p = constant Find the value of p that satises the boundary condition (|r| = R) = 0 (b) Find the surface charge density on the sphere as a function of cos = r z /R. (a) Image dipole: +q is displaced to +a, q is displaced to a. (r) = E0 z + kq (z + a)2 + x2 + y 2 kq = E0 z + 2 2za + a2 + x2 + y 2 2 + 2za + a2 + x2 + y 2 z z kq kq . = E0 z + r 1 (2za)/r2 + a2 /r2 r 1 + (2za)/r2 + a2 /r2 kq kq za za 1 + 2 + ... 1 2 + ... r r r r kq 2za = E0 z + + ... r r2 kp = E0 + 2 3 z. r kq (z a)2 + x2 + y 2 kq (6) (3) (4) (5)
(7)
Expand as a 0, qa = p: (r) = E0 z +
(8) (9)
Now impose the boundary condition (|r| = R) = 0: 0= E0 + 2 kp z R3 R r3
3
(10) (11) (12) R3 r3 (13)
kp = E0 R3 /2 (r) = E0 z 1
= E0 r cos 1
2 (b) Surface charge density: (R) = E(R) r /(4k) = |R r /(4k) 1 = 4k r R 2R3 E0 cos 1 + 3 4k r 3E0 cos = 4k p cos = 2R3 =
r=R
(14)
(15) (16)
2.2. Two metal spheres of radius a are inside a rectangular cavity in a grounded metal block, as shown in the gure. The metal spheres carry net charge +Q > 0 and Q.
(a) Sketch the electric eld lines in the cavity. Since the charges are equal and opposite on the two spheres, they must have the same number eld lines initiating or terminating on them. Make sure your eld lines are perpendicular to all metal surfacces they intersect. Draw a sucient number of eld lines such that there are eld lines initiating or terminating on all four surfaces of the rectangular cavity. Put arrows on your eld lines to indicate the direction of the electric eld. (b) For each eld line, put + or signs at the appropriate ends to indicate whether the local charge density is + or .
3 2.3. A metal sphere of radius a contains charge Q. (a) Find the electrostatic potential energy by integrating the energy density u(r) = over all space r > a. (b) If the charge Q is equal to the electron charge e = 1.6 1019 C, nd the radius r0 such that the total potential energy is equal to the electron rest mass energy me c2 = 0.511 KeV This is a typo: me c2 = 511. KeV. This is called the classical electron radius. To simplify the numerical calculation, you can use: ke2 = QED c = 197.3 KeV nm 137 (18) 12 E 8k (17)
This is another typo: c = 197.3 MeV fm = 0.1973 KeV nm. (a) Electrostatic energy: kq r2 1 (kq)2 u(r) = 8k r4 1 u(r) = 4k 2 q 2 8k r>R |E| = kq 2 1 = 2 r = (b) Electron radius: mc2 = R= = C = R= kq 2 2R kq 2 c kq 2 = 2 2mc c 2mc2 QED C /2 c 400fm mc2 400 fm 3 fm 137 kq 2 2R
R
r>R
r2 dr r4 (19) (20)
(21)
(22)
2.4. A charge +Q is at x = a, and a charge Q is at x = a. Evaluate the line integral dr E for the following cases. (a) From = A (0, 2a, 0) to B = (2a, 2a, 0) along the direction x. (b) From B to C = (2a, 0, 0) along the direction . y (c) From A to C along the circular arc of radius 2a. (d) From C to along the direction x. (23)
(e) From to A along the direction . y
(f) Show that the sum of the contours 2.4a + 2.4b is equal to the circular contour 2.4c.
(g) Show that the sum of the contours to A plus A to C plus C to is zero.
4 The electric eld: E(x, y, 0) = kQ (a) Integrate on curve A B:
B A (2a,2a,0)
[x2
x + (y a) x y + (y
3/2 a)2 ]
kQ
x + (y + a) x y [x2 + (y + a)2 ]
3/2
(24)
E dr =
Ex (x, y, z)dx
(0,2a,0) 2a
= kQ
0
dx
x [x2 + a2 ]3/2
x [x2 + 9a2 ]3/2
1/2 2a 0
= kQ x2 + a2
1/2
+ x2 + 9a2
1 1 1 1 = kQ + a a 5 a 13 3a (b) Integrate on curve B C:
C B (2a,0,0)
(25)
E dr = =
(2a,2a,0) 0 2a 2a
E dr
dyEy (2a, y, 0) dyEy (2a, y, 0)
0 2a
=
= kQ = kQ
dy
0 a a (4a2
[4a2 udu +
a
+ (y
ya
3/2 a)2 ]
y+a [4a2 + (y + a)2 ]
3/2 3/2
3a a
3/2 u2 )
+ kQ
udu (4a2 + u2 )
3a a
1 1 kQ = kQ 2 + u2 2 + u2 4a 4a a 1 1 . = kQ 0 + 13a 5a The change in potential from A to C is:
B C
(26)
C A =
A
E dr
B
E dr
1 1 1 1 1 1 + kQ + a a 5 a 13 3a 13a 5a kQ 2 = . a3 = kQ
(27)
5 (c) Integrate from A to C along arc dr = (2a)d:
C A 0
E dr =
/2
E 2ad
/2
(28) (29) [x2 x + (y a) x y + (y
3/2 a)2 ]
C A
= y cos x sin E dr = 2kQa x = 2a cos y = 2a sin
0
x + (y + a) x y [x2 + (y + a)2 ]
3/2
[ cos x sin ] d y
x2 + (y a)2 = (2a)2 4a2 sin + a2 x2 + (y + a)2 = 5a2 + 4a2 sin
C A /2
E dr = 2kQa2 u = sin
cos d a3 (5 4 sin ) +
3/2
+
cos d a3 (5 + r sin )
3/2
(30) (31)
0
du = cos d
1 0
C A
E dr =
2kQ a
du (5 4u)
3/2
du (5 + 4u)
3/2 1
1 1 kQ = 1/2 1/2 a (5 4u) (5 + 4u) 1 1 1 kQ 1 + = a 5 9 5 2kQ = 3a
0
(32)
Comparing the three previous results, note that the line integral from A to C is independent of the path. (d) In...