31. First, we need a formula for the field due to the arc. We use the notation for the charge density, = Q/L. Sample Problem 22-3 illustrates the simplest approach to circular arc field problems. Following the steps leading to Eq. 22-21, we see that the g
17. We make the assumption that bead 2 is in the lower half of the circle, partly because it would be awkward for bead 1 to slide through bead 2 if it were in the path of bead 1 (which is the upper half of the circle) and partly to eliminate a second solu
19. (a) Consider the figure below. The magnitude of the net electric field at point P is 1 q Enet = 2 E1 sin = 2 2 2 4 0 ( d / 2 ) + r
d /2
( d / 2)
2
+ r2
=
1
qd
4 0 ( d / 2 )2 + r 2 3/ 2
> For r > d , we write [(d/2)2 + r2]3/2 r3 so the expression abo
20. Referring to Eq. 22-6, we use the binomial expansion (see Appendix E) but keeping higher order terms than are shown in Eq. 22-7:
E=
q d 3 d2 1 d3 d 3 d2 1 d3 + 4 z2 + 2 z3 + 1 z + 4 z2 2 z3 + 2 1 + z 4o z
q d3 qd + + = 2o z3 4o z5
Therefore, in the t
21. Think of the quadrupole as composed of two dipoles, each with dipole moment of magnitude p = qd. The moments point in opposite directions and produce fields in opposite directions at points on the quadrupole axis. Consider the point P on the axis, a d
23. We use Eq. 22-3, assuming both charges are positive. At P, we have
Eleft ring = Eright ring 4 0 ( R + R
2
q1 R
2 3/ 2
)
=
q2 (2 R) 4 0 [(2 R ) 2 + R 2 ]3/ 2
Simplifying, we obtain
q1 2 = 2 q2 5
3/ 2
0.506.
25. From symmetry, we see that the net field at P is twice the field caused by the upper semicircular charge + q = R (and that it points downward). Adapting the steps leading to Eq. 22-21, we find
j Enet = 2
()
sin 4 0 R
90
90
q j. = 2 2 0 R
(a) With
26. We find the maximum by differentiating Eq. 22-16 and setting the result equal to zero.
d qz dz 4 z 2 + R 2 0
F GG H
c
I q R 2z J= h JK 4 cz + R h
2 3/ 2 0 2
2
2 5/ 2
=0
which leads to z = R / 2 . With R = 2.40 cm, we have z = 1.70 cm.
27. (a) The linear charge density is the charge per unit length of rod. Since the charge is uniformly distributed on the rod,
=
q 4.231015 C = = 5.19 1014 C/m. L 0.0815 m
(b) We position the x axis along the rod with the origin at the left end of the rod,
28. We use Eq. 22-16, with q denoting the charge on the larger ring:
qz qz 13 + = 0 q = Q 2 2 3/ 2 2 2 3/ 2 4 0 ( z + R ) 4 0 [ z + (3R) ] 5
3/ 2
= 4.19Q .
Note: we set z = 2R in the above calculation.
29. The smallest arc is of length L1 = r1 /2 = R/2; the middle-sized arc has length L2 = r2 / 2 = (2 R) / 2 = R ; and, the largest arc has L3 = (3R)/2. The charge per unit length for each arc is = q/L where each charge q is specified in the figure. Follow
30. (a) It is clear from symmetry (also from Eq. 22-16) that the field vanishes at the center. (b) The result (E = 0) for points infinitely far away can be reasoned directly from Eq. 2216 (it goes as 1/z as z ) or by recalling the starting point of its de
36. We write Eq. 22-26 as E z = 1 2 Emax ( z + R 2 )1/ 2 and note that this ratio is 2 (according to the graph shown in the figure) when z = 4.0 cm. Solving this for R we obtain R = z 3 = 6.9 cm.
1
8. (a) The individual magnitudes E1 and E2 are figured from Eq. 22-3, where the absolute value signs for q2 are unnecessary since this charge is positive. Whether we add the magnitudes or subtract them depends on if E1 is in the same, or opposite, directi
13. By symmetry we see the contributions from the two charges q1 = q2 = +e cancel each other, and we simply use Eq. 22-3 to compute magnitude of the field due to q3 = +2e. (a) The magnitude of the net electric field is
| Enet |= 1 2e 1 2e 1 4e = = 2 2 4 0
24. Studying Sample Problem 22-3, we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc is given by E=
sin 4 0 r
along the symmetry axis, with = q/r with in radians. In this problem, each charged quart
16. The net field components along the x and y axes are
Enet, x =
4 0 R
q1
2
q2 cos , 4 0 R 2
Enet, y =
q2 sin . 4 0 R 2
The magnitude is the square root of the sum of the components-squared. Setting the magnitude equal to E = 2.00 105 N/C, squaring and
15. (a) The electron ec is a distance r = z = 0.020 m away. Thus, (8.99 109 N m 2 C2 )(1.60 1019 C) = = 3.60 106 N/C . EC = 2 2 4 0 r (0.020 m)
e
(b) The horizontal components of the individual fields (due to the two es charges) cancel, and the vertical c
14. The field of each charge has magnitude E= kq e 1.60 1019 C =k = (8.99 109 N m 2 C2 ) = 3.6 106 N C. 2 2 2 r (0.020 m) ( 0.020 m )
The directions are indicated in standard format below. We use the magnitude-angle notation (convenient if one is using a
33. Consider an infinitesimal section of the rod of length dx, a distance x from the left end, as shown in the following diagram. It contains charge dq = dx and is a distance r from P. The magnitude of the field it produces at P is given by dE = 1 dx . 4
35. At a point on the axis of a uniformly charged disk a distance z above the center of the disk, the magnitude of the electric field is
E=
z 1 2 2 0 z + R2
LM N
OP Q
where R is the radius of the disk and is the surface charge density on the disk. See Eq
12. For it to be possible for the net field to vanish at some x > 0, the two individual fields (caused by q1 and q2) must point in opposite directions for x > 0. Given their locations in the figure, we conclude they are therefore oppositely charged. Furth
1. (a) We note that the electric field points leftward at both points. Using F = q0 E , and i orienting our x axis rightward (so points right in the figure), we find
N F = ( +1.6 1019 C ) 40 = (6.4 1018 N) i i C which means the magnitude of the force on t
2. We note that the symbol q2 is used in the problem statement to mean the absolute value of the negative charge which resides on the larger shell. The following sketch is for q1 = q2 .
The following two sketches are for the cases q1 > q2 (left figure) an
3. Since the magnitude of the electric field produced by a point charge q is given by E =| q | / 4 0 r 2 , where r is the distance from the charge to the point where the field has magnitude E, the magnitude of the charge is
q
( 0.50 m ) ( 2.0 N C ) = 5.6
5. Since the charge is uniformly distributed throughout a sphere, the electric field at the surface is exactly the same as it would be if the charge were all at the center. That is, the magnitude of the field is q E= 4 0 R 2 where q is the magnitude of th
7. At points between the charges, the individual electric fields are in the same direction and do not cancel. Since charge q2= 4.00 q1 located at x2 = 70 cm has a greater magnitude than q1 = 2.1 108 C located at x1 = 20 cm, a point of zero field must be c
9. The x component of the electric field at the center of the square is given by
Ex = | q1 | | q3 | | q2 | | q4 | cos 45 + 2 2 2 2 (a / 2) (a / 2) (a / 2) (a / 2) 1 1 1 = (| q1 | + | q2 | | q3 | | q4 |) 2 4 0 a / 2 2 1 4 0 = 0.
Similarly, the y component