Physics 2102 Gabriela Gonzlez
Carl Friedrich Gauss 1777-1855
=
Electric Flux A surface integral!
E dA
CLOSED surfaces: define the vector dA as pointing OUTWARDS Inward E gives negative Outward E gives positive
1
Consider any ARBITRARY CLOSED surface
49. At all points where there is an electric field, it is radially outward. For each part of the problem, use a Gaussian surface in the form of a sphere that is concentric with the sphere of charge and passes through the point where the electric field is
50. (a) Eq. 25-22 yields
U= 1 1 CV 2 = 200 1012 F 7.0 103 V 2 2
c
hc
h
2
= 4.9 103 J .
(b) Our result from part (a) is much less than the required 150 mJ, so such a spark should not have set off an explosion.
5. The magnitude of the force of either of the charges on the other is given by
F=
b
1 q Qq
r2
4 0
g
where r is the distance between the charges. We want the value of q that maximizes the
function f(q) = q(Q q). Setting the derivative df/dq equal to zero
69. We apply conservation of energy for the particle with q = 7.5 106 C (which has zero initial kinetic energy):
U0 = Kf + Uf
where U =
qQ . 4or
(a) The initial value of r is 0.60 m and the final value is (0.6 + 0.4) m = 1.0 m (since the particles repel e
37. We combine Eq. 22-9 and Eq. 22-28 (in absolute values). F= qE= q
FG p IJ = 2kep H 2 z K z
3 3 0
where we have used Eq. 21-5 for the constant k in the last step. Thus, we obtain
F= 2 8.99 10 9 N m 2 C 2 1.60 10 19 C 3.6 10 29 C m
c
hc
hc
h
c25 10 mh
9
48. Let EA designate the magnitude of the field at r = 2.4 cm. Thus EA = 2.0 107 N/C, and is totally due to the particle. Since Eparticle = q 4o r2
then the field due to the particle at any other point will relate to EA by a ratio of distances squared. No
45. (a) The electric field in the region between the plates is given by E = V/d, where V is the potential difference between the plates and d is the plate separation. The capacitance is given by C = 0A/d, where A is the plate area and is the dielectric co
70. (a) Using d = 2 m, we find the potential at P:
VP =
1 +2e 1 -2e 1e + = . 40 d 40 2d 40 d
Thus, with e = 1.60 1019 C, we find VP = 7.19 1010 V. Note that we are implicitly assuming that V 0 as r . (b) Since U = qV , then the movable particle's contribu
41. (a) The magnitude of the force acting on the proton is F = eE, where E is the magnitude of the electric field. According to Newtons second law, the acceleration of the proton is a = F/m = eE/m, where m is the mass of the proton. Thus,
. c160 10 Chc2.0
52. Initially the capacitors C1, C2, and C3 form a series combination equivalent to a single capacitor which we denote C123. Solving the equation 1 1 1 C1 + C 2 + C3 = 1 C123 ,
we obtain C123 = 2.40 F. With V = 12.0 V, we then obtain q = C123V = 28.8 C. I
8. (a) The individual force magnitudes (acting on Q) are, by Eq. 21-1,
k
q1 Q
ba g
a2
2
=k
q2 Q
ba g
a2
2
which leads to |q1| = 9.0 |q2|. Since Q is located between q1 and q2, we conclude q1 and q2
are like-sign. Consequently, q1/q2 = 9.0.
(b) Now we have
71. The derivation is shown in the book (Eq. 24-33 through Eq. 24-35) except for the change in the lower limit of integration (which is now x = D instead of x = 0). The result is therefore (cf. Eq. 24-35)
L + L2 + d2 V= ln 4o D + D2 + d2
2.0 x 10-6 4 + 1
50. The field is zero for 0 r a as a result of Eq. 23-16. Thus, (a) E = 0 at r = 0, (b) E = 0 at r = a/2.00, and (c) E = 0 at r = a. For a r b the enclosed charge qenc (for a r b) is related to the volume by qenc = Therefore, the electric field is 1 qenc
40. (a) The initial direction of motion is taken to be the +x direction (this is also the direction of E ). We use v 2 vi2 = 2ax with vf = 0 and a = F m = eE me to solve for f distance x:
31 6 vi2 me vi2 9.11 10 kg 5.00 10 m s x = = = = 7.12 10 2 m. 3 19
51.One way to approach this is to note that since they are identical the voltage is evenly divided between them. That is, the voltage across each capacitor is V = (10/n) volt. With C = 2.0 106 F, the electric energy stored by each capacitor is 1 CV2. The
4. The fact that the spheres are identical allows us to conclude that when two spheres are
in contact, they share equal charge. Therefore, when a charged sphere (q) touches an
uncharged one, they will (fairly quickly) each attain half that charge (q/2). W
49. (a) Initially, the capacitance is
C0 =
0A
d
d8.85 10 =
(8.85 10
12
12 10 .
C2 N m2 2
. i (012 m ) = 89 pF.
2
m
(b) Working through Sample Problem 25-7 algebraically, we find:
C=
0 A ( d b) + b
12
=
(4.8)(1.2 0.40)(10 m) + (4.0 103 m)
C2 N m 2 2
) (0.
68. The escape speed may be calculated from the requirement that the initial kinetic energy (of launch) be equal to the absolute value of the initial potential energy (compare with the gravitational case in chapter 14). Thus, 1 eq 2 2 m v = 4o r where m =
44. (a) The flux is still 750 N m 2 /C , since it depends only on the amount of charge enclosed. (b) We use = q / 0 to obtain the charge q: q = 0 = 8.85 10
12
C2 N m2
( 750 N m
2
/ C = 6.64 109 C.
)
65. We treat the system as a superposition of a disk of surface charge density and radius R and a smaller, oppositely charged, disk of surface charge density and radius r. For each of these, Eq 24-37 applies (for z > 0)
V=
2 0
e
z2 + R2 z +
j
2 0
e
z2 +
1. (a) With a understood to mean the magnitude of acceleration, Newton's second and third laws lead to m2 a2 = m1a1
c6.3 10 kghc7.0 m s h = 4.9 10 m =
-7 2 2
-7
9.0 m s
2
kg.
(b) The magnitude of the (only) force on particle 1 is
q q q F = m1a1 = k 1 2 2
34. (a) Vertical equilibrium of forces leads to the equality
q E = mg E= mg . 2e
Using the mass given in the problem, we obtain E = 2.03 107 N C . (b) Since the force of gravity is downward, then qE must point upward. Since q > 0 in this situation, this i
46. (a) The electric field E1 in the free space between the two plates is E1 = q/0A while that inside the slab is E2 = E1/ = q/0A. Thus, V0 = E1 d b + E2b = and the capacitance is 8.85 1012 NCm2 (115 104 m 2 ) ( 2.61) 0 A q C= = = = 13.4pF. V0 ( d b ) + b
45. (a) Since r1 = 10.0 cm < r = 12.0 cm <r2 = 15.0 cm,
8.99 109 N m 2 C2 4.00 108 C 1 q1 E (r ) = = = 2.50 104 N/C. 2 2 4 0 r ( 0.120 m )
(
)(
)
(b) Since r1 < r2 < r = 20.0 cm, 8.99 109 N m 2 / C2 ( 4.00 + 2.00 ) 1108 C 1 q1 + q2 E (r ) = = = 1.35 104 N
66. Since the electric potential energy is not changed by the introduction of the third particle, we conclude that the net electric potential evaluated at P caused by the original two particles must be zero: q1 q2 + =0. 4o r1 4o r2 Setting r1 = 5d/2 and r