A spherical ball of charged particles has a uniform charge density. In terms of the ball's radius R, at what radial distances in
side and outside the ball is the magnitude of the ball's electric field equal to 1 3 of the maximum magnitude of that field? (a) inside the ball R (b) outside the ball R
1 answer:
Answer:
Electric field at radius r inside the solid sphere is
Electric field at radius r between inner radius and outer radius of the shell is
E=0 N/C
Explanation:Given
The radius of the solid sphere is
The charge on the solid sphere is
The inner radius of the shell is
The outer radius of the shell is
The total charge on the shell is
PART(A)
The magnitude of electric field at radius r where \\The volumetric charge density of the solid sphere will be
The charge enclosed by the radius r inside the solid sphere is
rho=
According to gauss law
PART(B)
The electric field at radius r where
The shell is conducting so due to induction of charge
The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell
So,
The charge on the inner surface of the shell is
Due to conservation of the charge on the shell
The charge accumulated on the outer surface of the shell is
The charge enclosed by the radius r where
According to gauss law
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The complete text of the problem is:
<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>
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<u>Solution:</u>
1) Acceleration: on the hardwood floor,
on the carpeted floor
First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation
where
v is the final speed
u = 0 is the initial speed (the child starts from rest)
a = g = 9.8 m/s^2 is the acceleration of gravity
d = 0.43 m is the distance covered by the child as he falls from the bed
Solving for v,
Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is
So we can find the acceleration by using again the equation
Solving for a,
For the carpeted floor instead,
therefore the acceleration is
2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor
We can find the duration of the collision in both cases by using the equation of the acceleration
where
v' = 0
v = 2.9 m/s
For the hardwood floor,
So the duration of the collision is
For the carpeted floor,
So the duration of the collision is
We can now comment the results using the initial statement of the problem:
"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"
Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).