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There are two concentric metal shells of radii r1 and r2 (<r1). If the outer shell has a charge q and the inner shell is grounded, the charge on the inner shell is :
- a)zero
- b)-(r1/r2)q
- c)r1r2q
- d)∞
- e)None of these
Correct answer is option 'E'. Can you explain this answer?
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There are two concentric metal shells of radii r1 and r2 (<r1). If ...
With r2 > r1) and thickness t. The inner shell has a charge of Q.
a) Find the electric field inside the inner shell.
Since the charge is distributed on the inner surface of the shell, the electric field inside the shell is zero. This is because any electric field that would be present would cause the charges to move, but they are constrained to stay on the inner surface of the shell.
b) Find the electric field between the two shells.
Using Gauss's law, we can find the electric field between the two shells by considering a Gaussian surface that is a cylinder with its axis along the radial direction, extending from the inner radius to the outer radius of the shells. The electric flux through this surface is given by:
Φ = E * A
where E is the electric field, and A is the surface area of the Gaussian surface. Since the electric field is constant and parallel to the surface of the cylinder, the flux is simply:
Φ = E * A = E * 2πrL
where L is the length of the cylinder. The charge enclosed by the Gaussian surface is Q, so Gauss's law gives:
Φ = Q / ε0
where ε0 is the permittivity of free space. Solving for E, we get:
E = Q / (2πε0Lr)
c) Find the electric field outside the outer shell.
Again using Gauss's law, we can consider a Gaussian surface that is a sphere with radius r > r2. The electric flux through this surface is given by:
Φ = E * A = E * 4πr^2
Since there is no charge enclosed by this surface, Gauss's law gives:
Φ = 0
Therefore, the electric field outside the outer shell is zero. This is because the charges on the inner shell are shielded by the outer shell, which acts like a Faraday cage. Any electric field that would be present outside the outer shell is cancelled out by the charges on the outer surface of the shell.
a) Find the electric field inside the inner shell.
Since the charge is distributed on the inner surface of the shell, the electric field inside the shell is zero. This is because any electric field that would be present would cause the charges to move, but they are constrained to stay on the inner surface of the shell.
b) Find the electric field between the two shells.
Using Gauss's law, we can find the electric field between the two shells by considering a Gaussian surface that is a cylinder with its axis along the radial direction, extending from the inner radius to the outer radius of the shells. The electric flux through this surface is given by:
Φ = E * A
where E is the electric field, and A is the surface area of the Gaussian surface. Since the electric field is constant and parallel to the surface of the cylinder, the flux is simply:
Φ = E * A = E * 2πrL
where L is the length of the cylinder. The charge enclosed by the Gaussian surface is Q, so Gauss's law gives:
Φ = Q / ε0
where ε0 is the permittivity of free space. Solving for E, we get:
E = Q / (2πε0Lr)
c) Find the electric field outside the outer shell.
Again using Gauss's law, we can consider a Gaussian surface that is a sphere with radius r > r2. The electric flux through this surface is given by:
Φ = E * A = E * 4πr^2
Since there is no charge enclosed by this surface, Gauss's law gives:
Φ = 0
Therefore, the electric field outside the outer shell is zero. This is because the charges on the inner shell are shielded by the outer shell, which acts like a Faraday cage. Any electric field that would be present outside the outer shell is cancelled out by the charges on the outer surface of the shell.
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There are two concentric metal shells of radii r1 and r2 (<r1). If the outer shell has a charge q and the inner shell is grounded, the charge on the inner shell is :a)zerob)-(r1/r2)qc)r1r2qd)∞e)None of theseCorrect answer is option 'E'. Can you explain this answer?
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