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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Hello
Let's solve the problem in the three different steps
1) Uniformly accelerated motion, with acceleration
and for a total time of 
. The body is initially at rest, so the distance covered is given by
Calling
and 
the final and initial velocity, and since the 
because the body starts from rest, we can use
to find the final velocity after this first leg:
And the average velocity in this first leg is
2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg:
. This is also the average velocity of the second leg.
The total time of this second leg is
. The distance covered is given by
3) Uniformly decelerated motion, with constant deceleration of
and for a total time of 
. Here, the initial velocity of the body is the final velocity of the previous leg, i.e. 
. Therefore, the distance covered in this leg is given by
The final velocity in this leg is given by
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
4) Finally, the total distance covered in the motion is
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
Let's solve the problem in the three different steps
1) Uniformly accelerated motion, with acceleration
Calling
to find the final velocity after this first leg:
And the average velocity in this first leg is
2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg:
The total time of this second leg is
3) Uniformly decelerated motion, with constant deceleration of
The final velocity in this leg is given by
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
4) Finally, the total distance covered in the motion is
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg: