A question about conducting sphere using Gauss's Law

AI Thread Summary
To find the surface charge density on the inner and outer surfaces of a conducting spherical shell with a point charge at its center, the inner surface will have a charge of -Q due to the positive point charge, while the outer surface will have a charge of -2Q, resulting from the total charge of -3Q on the shell. The surface charge density can be calculated using the formula σ = Q/A, where A is the area of the spherical surfaces. The area of a sphere is given by A = 4πr², where r is the radius of the respective surface. The inner surface area is based on radius a, and the outer surface area is based on radius b. Understanding these areas is crucial for calculating the surface charge densities accurately.
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Homework Statement



A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q located at its center. The total charge on the shell is -3Q, and it is insulated form it surroundings. What is the surface charge density on the inner surface of the conducting shell.? What is the surface charge density on the outer surface?

Homework Equations



EA= Q enclosed / ε0

The Attempt at a Solution



I know that since there is a point charge, this will change up the configuration a bit making the inner surface having -Q and outer surface to have -2Q charge enclosed. I have problems in determining the area. I don't know when does the inner end and when does the outer surface begin. Is it half way?? What would be the area then?
 
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The radii of the inner and outer surfaces are given, a and b. Do you know how to calculate the area of a spherical surface from the radius?

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