Modification to metal sphere problem

  • Thread starter Thread starter cherrios
  • Start date Start date
  • Tags Tags
    Metal Sphere
AI Thread Summary
The discussion revolves around a problem involving a metal sphere with a radius of x, surrounded by a conducting shell with inner radius y and outer radius Z, carrying a total charge of 4 microcoulombs. The task is to find the electric field between the outer surface of the metal sphere and the inner radius of the conducting shell. A Gaussian surface is necessary for this calculation, as it helps apply Gauss's law effectively. The inner sphere has a charge of 1 microcoulomb, and the electric field can be determined using the formula derived from Gauss's law. The outer shell does not contribute to the electric field in this region due to the absence of charge between the inner and outer surfaces.
cherrios
Messages
8
Reaction score
0
This problem is somewhat similar to the one I had posted yesterday.

There is a metal spehre, radius=x, that it surrounded by a conducting shell (also spherical) that has an inner radius=y and outer radius=Z. the total charge is 4 micro Coloumbs--> 1 micro Coloumb on the inner sphere, and the rest is distributed in the shell

1)Find electric field between the outer surface of the metal sphere and the inner radius of the conducting shell.

Would I need to take a Gaussian surface between the outer surface of the metal sphere and inner radius of the conducting shell? And also, how would I find the surface charge density on the inner and outer surfaces of the conducting shell?
 
Physics news on Phys.org
Assuming that you have no induction and uniform chage density, the outer shell will not contribute to the electric field, since it has no charge between it and inner shell. You simply have q_{in} = 1\mu C, and by Gauss's law for a spherical shell
\int \vec{E}.d\vec{A} = \int EdA = E \int dA = E(4 \pi r^2) = \frac{q_{in}}{\epsilon_0}, which will yield E(r).
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top