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.
1 answer:
Answer:
The specific question is not stated, however the general idea is given in the attached picture. The electric field in each region can be found by Gauss’ Law.
at r < R:
Since the solid sphere is conducting, the total charge Q is distributed over the surface, and the electric field inside the sphere is zero.
E = 0.
at R < r < 2R:
The electric field can be found by Gauss’ Law as in the attachment. The green pencil shows this exact region.
at 2R < r:
The electric field can again be found by Gauss’ Law, the blue pencil shows the calculations for this region.
Explanation:
Gauss’ Law is straightforward when applied to spheres. The area of the sphere is , and the enclosed charge is given in the question as Q for the inner sphere, and 2Q for the whole system.
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Answer:
h2 = 0.092m
Explanation:
From a balance of energy from point A to point B, we get speed before the collision:
Solving for Vb:
Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:
Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:
Solving for h2:
h2 = 0.092m