CP In experiments in which atomic nuclei collide, head-on collisions like that described in Problem 23.74 do happen, but “near misses” are more common. Suppose the alpha particle in that problem is not “aimed” at the center of the lead nucleus but has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L = p 0 b, where p 0 is the magnitude of the particle’s initial momentum and b = 1.00 × 10− 12 m. What is the distance of closest approach? Repeat for b = 1.00 × 10− 12 m and b = 1.00 × 10− 14 m.
CP In experiments in which atomic nuclei collide, head-on collisions like that described in Problem 23.74 do happen, but “near misses” are more common. Suppose the alpha particle in that problem is not “aimed” at the center of the lead nucleus but has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L = p 0 b, where p 0 is the magnitude of the particle’s initial momentum and b = 1.00 × 10− 12 m. What is the distance of closest approach? Repeat for b = 1.00 × 10− 12 m and b = 1.00 × 10− 14 m.
CP In experiments in which atomic nuclei collide, head-on collisions like that described in Problem 23.74 do happen, but “near misses” are more common. Suppose the alpha particle in that problem is not “aimed” at the center of the lead nucleus but has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L = p0b, where p0 is the magnitude of the particle’s initial momentum and b = 1.00 × 10−12 m. What is the distance of closest approach? Repeat for b = 1.00 × 10−12 m and b = 1.00 × 10−14 m.
Definition Definition Product of the moment of inertia and angular velocity of the rotating body: (L) = Iω Angular momentum is a vector quantity, and it has both magnitude and direction. The magnitude of angular momentum is represented by the length of the vector, and the direction is the same as the direction of angular velocity.
1. A charge of -25 μC is distributed uniformly throughout a spherical volume of radius 11.5 cm.
Determine the electric field due to this charge at a distance of (a) 2 cm, (b) 4.6 cm, and (c) 25 cm from
the center of the sphere.
(a) =
=
(b) E =
(c)Ẻ =
=
NC NC NC
1.
A long silver rod of radius 3.5 cm has a charge of -3.9
ис
on its surface. Here ŕ is a unit vector
ст
directed perpendicularly away from the axis of the rod as shown in the figure.
(a) Find the electric field at a point 5 cm from the center of the rod (an outside point).
E =
N
C
(b) Find the electric field at a point 1.8 cm from the center of the rod (an inside point)
E=0
Think & Prepare
N
C
1. Is there a symmetry in the charge distribution? What kind of symmetry?
2. The problem gives the charge per unit length 1. How do you figure out the surface charge density σ
from a?
1. Determine the electric flux through each surface whose cross-section is shown below.
55
S₂
-29
S5
SA
S3
+ 9
Enter your answer in terms of q and ε
Φ
(a) s₁
(b) s₂
=
-29
(C) Φ
զ
Ερ
(d) SA
=
(e) $5
(f) Sa
$6
=
II
✓
-29
S6
+39
Chapter 23 Solutions
University Physics with Modern Physics (14th Edition)
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