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Solved: Concentric Spherical Shells. A small conducting
Chapter 22, Problem 22.47(choose chapter or problem)
Concentric Spherical Shells. A small conducting spherical shell with inner radius and outer radius is concentric with a larger conducting spherical shell with inner radius and outer radius (Fig. P22.47). The inner shell has total charge and the outer shell has charge (a) Calculate the electric field (magnitude and direction) in terms of and the distance from the common center of the two shells for (i) (ii) (iii) (iv) (v) Show your results in a graph of the radial component of as a function of (b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (iii) inner surface of the large shell; (iv) outer surface of the large shell?
Questions & Answers
QUESTION:
Concentric Spherical Shells. A small conducting spherical shell with inner radius and outer radius is concentric with a larger conducting spherical shell with inner radius and outer radius (Fig. P22.47). The inner shell has total charge and the outer shell has charge (a) Calculate the electric field (magnitude and direction) in terms of and the distance from the common center of the two shells for (i) (ii) (iii) (iv) (v) Show your results in a graph of the radial component of as a function of (b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (iii) inner surface of the large shell; (iv) outer surface of the large shell?
ANSWER:Step 1 of 5
Given: A small conducting spherical shell with inner radius and outer radius as shown in figure.
These are usually difficult to do but they can usually be done with gauss’s law,
So, if we put a spherical surface inside the inner surface of the sphere, we will get 0 flux so that means there is 0 charge inside the first sphere which is obvious. If we put a spherical surface between the inner surface and these are static charges so again the inner surface must have a charge of 0. Since the inner surface of the inner surface has a charge of 0 that means the outer surface of the inner sphere must have a charge of +2q because the sphere itself must have a charge +2q.
Now we have the inner sphere complete.