CALC A Nonuniformly Charged Slab. Repeat Problem 22.54, but now let the charge density of the slab be given by ρ ( x ) = ρ 0 ( x / d ) 2 , where ρ 0 is a positive constant. 22.54 A Uniformly Charged Slab. A slab of insulating material has thickness 2 d and is oriented so that its faces are parallel to the yz -plane and given by the planes x = d and x = − d . The y - and z -dimensions of the slab are very large compared to d ; treat (hem as essentially infinite. The slab has a uniform positive charge density ρ . (a) Explain why the electric field due to the slab is zero at the center of the slab ( x = 0). (b) Using Gauss’s law, find the electric field due to the slab (magnitude and direction) at all points in space.
CALC A Nonuniformly Charged Slab. Repeat Problem 22.54, but now let the charge density of the slab be given by ρ ( x ) = ρ 0 ( x / d ) 2 , where ρ 0 is a positive constant. 22.54 A Uniformly Charged Slab. A slab of insulating material has thickness 2 d and is oriented so that its faces are parallel to the yz -plane and given by the planes x = d and x = − d . The y - and z -dimensions of the slab are very large compared to d ; treat (hem as essentially infinite. The slab has a uniform positive charge density ρ . (a) Explain why the electric field due to the slab is zero at the center of the slab ( x = 0). (b) Using Gauss’s law, find the electric field due to the slab (magnitude and direction) at all points in space.
CALC A Nonuniformly Charged Slab. Repeat Problem 22.54, but now let the charge density of the slab be given by ρ(x) = ρ0(x/d)2, where ρ0 is a positive constant.
22.54 A Uniformly Charged Slab. A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes x = d and x = −d. The y- and z-dimensions of the slab are very large compared to d; treat (hem as essentially infinite. The slab has a uniform positive charge density ρ. (a) Explain why the electric field due to the slab is zero at the center of the slab (x = 0). (b) Using Gauss’s law, find the electric field due to the slab (magnitude and direction) at all points in space.
A block of mass m₁
=
10.0 kg is connected to a block of mass m₂
34.0 kg by a massless string that passes over a light, frictionless pulley. The 34.0-kg block is connected to a spring that has negligible mass and a force constant of k = 200 N/m as shown in the figure below. The spring is
unstretched when the system is as shown in the figure, and the incline is frictionless. The 10.0-kg block is pulled a distance h = 22.0 cm down the incline of angle = 40.0° and released from rest. Find the speed of each block when the spring is again unstretched.
Vm1
×
1.32
Vm2
= 1.32
×
m/s
m/s
A block of mass m₁ = 10.0 kg is connected to a block of mass m₂ = 34.0 kg by a massless string that passes over a light, frictionless pulley. The 34.0-kg block is connected to a spring that has negligible mass and a force constant of k = 200 N/m as shown in the figure below. The spring is
unstretched when the system is as shown in the figure, and the incline is frictionless. The 10.0-kg block is pulled a distance h = 22.0 cm down the incline of angle 0 = 40.0° and released from rest. Find the speed of each block when the spring is again unstretched.
m/s
Vm1
Vm2
m/s
mi
m2
k
i
Truck suspensions often have "helper springs" that engage at high loads. One such arrangement is a leaf spring with a helper coil spring mounted on the axle, as in the figure below. The helper spring engages when the main leaf spring is compressed by distance yo, and then helps to
support any additional load. Consider a leaf spring constant of 5.45 × 105 N/m, helper spring constant of 3.60 × 105 N/m, and y = 0.500 m.
Truck body
Dyo
Axle
(a) What is the compression of the leaf spring for a load of 4.90 × 105 N?
m
(b) How much work is done compressing the springs?
]
Chapter 22 Solutions
University Physics with Modern Physics Plus Mastering Physics with eText -- Access Card Package (14th Edition)
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