A solid insulating sphere of radius R = 1.0 m that carries a positive charge Q1 = 1.0 mC uniformly distributed over it is concen
tric with a very thin insulating shell of radius 2R that carries the charge Q2 = -1.0 mC. The charge Q2 is distributed uniformly over the shell.
Use Gauss's law to obtain the magnitude of the electric field:
a. E1 inside the sphere, r < R
b. E2 between the sphere and the shell, R < r < 2R
c. E3 outside the shell, r > 2R
d. Plot the electric field in the radial direction on a graph. Be sure to label the actual magnitude of the electric field at r = 0, r= R, and r = 2R.
Use Gauss's law to obtain the magnitude of the electric field:
a. E1 inside the sphere, r < R
b. E2 between the sphere and the shell, R < r < 2R
c. E3 outside the shell, r > 2R
d. Plot the electric field in the radial direction on a graph. Be sure to label the actual magnitude of the electric field at r = 0, r= R, and r = 2R.
1 answer:
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Answer:
The potential for r > rb is equal to zero.
Explanation:
For r > rb, the potential is:
Then, the net potential is:
As we know that centripetal acceleration is given as
here we know that
linear speed = 35 m/s
here radius of path is same as length of arm
R = 0.7 m
now from above above formula we have
so acceleration is 1750 m/s/s