A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius 2R that also carries charge Q. The charge Q is distributed uniformly over the insulating shell. (a) Find the electric field (magnitude and direction) in each of the regions 0 < r < R, R < r < 2R, and r > 2R. (b) Graph the electric-field magnitude as a function of r.
Solution 45P Step 1 of 4: Given solid conducting sphere of radius R which carries a charge Q and has a concentric very thin insulating shell of radius 2R that also carries a charge Q uniformly distributed over insulating shell as shown in the figure below. Here we need to calculate the in part(a) the electric field magnitude and direction in terms of r from the center in different regions using Gauss’s law, whereas in part(b) we need to plot the graph of electric field as a function of r using results of part(a). Step 2 of 4 (a) Find the electric field (magnitude and direction) in each of the regions 0 < r < R, R < r < 2R, and r > 2R. The regions can be represented on the figure as shown below, (i) For region 0 < r < R, since the region lies inside the conductor,as the electric field inside the conductor is zero. Hence electric field in this region will be zero. That is E=0 for 0 < r < R (ii) For R < r < 2R , by drawing gaussian surface at this region with radius r , the charge 2 enclosed will be Q and the area will be A = 4r and hence the electric field using Gauss’s law will be, E= 1 Q2 radially outward for R < r < 2R 40 r