anonymous
  • anonymous
A thin metallic spherical shell contains a charge Q on it. A point charge q is placed at the center of the shell and another charge q1 is places outside it. All the three charges are positive. The force on the charge at the center is: A.) towards left B.) towards right C.) upward D.) zero. The answer is given as D. How is this so? Won't the charge q outside the shell exert a force on the charge in the center?
OCW Scholar - Physics II: Electricity and Magnetism
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Please view the figure attached.
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anonymous
  • anonymous
Here's the continuation of the above problem: The force on the central charge is: A.) towards left B.) towards right C.) upward D.) zero. The answer is given as B. ?????? Isnt' the field inside a charged metallic (conducting) shell zero.
anonymous
  • anonymous
The central charge is shielded from the effect of the outer charge by the conducting shell. Think of the Faraday Cage. Hence, no force on it. Just use Gauss's law. You'll find that the only field inside the conducting shell is due to the inner charge, hence it will not feel the outer charge.

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anonymous
  • anonymous
Correction to the second part of the question: The force on the central charge DUE TO THE SHELL is: A.) towards left B.) towards right C.) upward D.) zero. The answer is given as B. ?????? Isn't' the field inside a charged metallic (conducting) shell zero.
anonymous
  • anonymous
Let's see. Since there is a charge inside the conducting shell, there should be some negative charge -q1 distributed in the inside of the shell, because there can't be an electric field inside the shell itself. There is complete symmetry, so there shouldn't be any force on the inside charge due to the shell, even though the central charge is in unstable equilibrium. So if the system is dynamic, which means that the outer charge will repel the shell, pushing it to the left. Then, the center charge would be attracted to the right side of the shell.
anonymous
  • anonymous
Sorry, it should have been "charge -q" no -q1

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