Saturday, March 20, 2010

Irodov Problem 3.17

a) First let us try and understand what happens in this problem. As shown in the figure, two charged spheres with charge densities are and respectively are placed on top of each other displaced by a small distance of . While the -ive and the +ive charges from the two spheres cancel in the center of the spheres. However, since the spheres are displaced with respect to each other (the upper sphere is +ively charged) around the periphery the charges will not cancel each other as shown in the figure. The amount of charge in the periphery will however, depend on the polar angle . As seen from the figure, the charge will be zero at and the maximum at . In fact what Irodov asks to prove is that is the displacement is small enough, then the two spheres can be replaced by a single hollow sphere with a surface change density that depends on the angle as .

To prove this, let us first determine the thickness of the charged section as a function of the polar angle . As seen from the figure,







Let us now consider an infinitesimally small piece of the charged part of the sphere that has the polar and azimuthal angles of and respectively and subtends a polar angle of and an azimuthal angle of as shown in the figure. The surface area of this section will be,



The charge contained in this section will this be,



If there were a single hollow sphere with a surface charge density then we would have,



Comparing (3) and (4) we have,







b) Let us assume that the center of the positively charged sphere is located at the origin. The electric field inside the positively charged sphere distance r is given by,








Since the center of the negatively charged sphere is located units below that of the positively charged sphere, the electric field due to this sphere is given by,





The net electric field due to the two sphere are given by,

No comments:

Post a Comment