8. Consider two nested, spherical conducting shells. The first has inner radius

Question

8. Consider two nested, spherical conducting shells. The first has inner radius a and outer radius b. The second has inner radius e and outer radius d. The system is shown in Fig. 5. In the following four situations, determine the total charge on each of the faces of the conducting spheres (inner and outer for each), as wel as the electric field and potential everywhere in space (as a function of distance r from the center of the spherical shells). In all cases the shells begin uncharged, and a charge is then instantly introduced somewhere. (i) Both shells are not connected to any other conductors (floating) that is, their net charge will remain fixed. A positive charge +Q is introduced into the center of the inner spherical shell. Take the zero of potential to be at infinity. (ii) The inner shell is not connected to ground (floating) but the outer shel is grounded that is, it is fixed at V- 0 and has whatever charge is necessary on it to maintain this potential. A negative charge -Q is introduced into the center of the inner spherical she. () The inner shel is grounded but the outer shel is floating. A positive charge is introduced into the center of the inner spherical shell. (iv) Finally, the outer shell is grounded and the inner she is floating. This time the positive charge +Q is introduced into the region in between the two shells. In this case the question "What are E( and V(r)? cannot be answered analytically in some regions of space. In the regions where these questions can be answered analytically, give answers. In the regions where they cannot be answered analytically, expl why, but try to draw what you think the electric field should look like and give as much information about the potential as possible.

Explanation / Answer

We know that charge resides on the surface of a conductor. So for inner sphere we have

(i) when a +Q charge is placed inside the inner sphere, it will induce some charge on the inner shell wall (a). To determine the induced charge we will apply Gauss's Law on a spherical surface S. Electrostatic field E is zero at each point on the surface S and total charge on the surface S is also zero. So, induced charge at a is -Q and induced charge at b is Q (total charge of the hollow sphere must be zero). For the outer sphere, it will experience Q charge because of the surface charge of the inner shell. So charge induced at c is -Q and at d is +Q. As total charge of the outer hollow sphere is also zero.

(ii) Due to -Q charge inside the inner sphere, total charge on a is +Q and total charge on d is -Q. This is because electrostatic field at each point of the sphere is zero and total charge of the sphere is also zero.

In a grounded conductor, infinite amount of charges flow to or from ground to keep the surface potential zero. So we have to consider image charge to find the total charge on such a surface.

Therefore, for the outer grounded sphere, charge at c is zero and at d is Q.

(iii) As the inner sphere is grounded and a +Q charge placed inside it, charge at a is zero, charge at b is -Q. For the outer sphere, it will feel a -Q charge at it's centre. So, charge at c is +Q and at d is -Q.

(iv) As I have already answered 12 parts of the question, I am skipping this one according to chegg's policy.

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