HIII DROPPING COINS YALL
2 answers:
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Answer:
at r < R;
at 2R > r > R;
at r >= 2R
Explanation:
Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,
(integral over a closed surface)
where,
= Electric field
= charged enclosed within the closed surface
= permittivity of free space
Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .
at r < R,
= 0 and hence
= 0 (since the sphere is conducting, all the charges get repelled towards the surface)
at 2R > r > R,
= Q,
therefore,
(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to )
or,
at r >= 2R
= 2Q
Hence, by similar calculations, we get,