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3 years ago

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A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius

2R that also carries charge Q. The charge Q is distributed uniformly over the insulating shell.
(a) Find the electric field (magnitude and direction) in each of the regions 0 < r < R, R < r < 2R, and r > 2R.
(b) Graph the electric-field magnitude as a function of r.

1 answer:

chubhunter [2.5K]3 years ago

7 0

Answer:

a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2

b) See the picture

Explanation:

We can use Gauss's law to find the electric field in all the regions:

EA = qen/e0 where qen is the enclosed charge

Remember that the electric field everywhere outside a sphere is:

E(r) = q/(4*pi*eo*r^2) = Kq/r^2

a)

For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0 EA = 0/e0 = 0 E = 0
For R < r < 2R: Here the enclosed charge is equal Q E = Q/(4*pi*eo*r^2) = KQ/r^2
For r > 2R: Here the enclosed charge is equal 2Q E = Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2

b) At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance

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