Two charged concentric spherical shells have radii of 10.5 cm and 14.0 cm. Thech

Question

Two charged concentric spherical shells have radii of 10.5 cm and 14.0 cm. Thecharge on the inner shell is 4.50 x10^-8 C and that on the outer shell is 2.40 x10^-8 C. Find the magnitude of the electricfield at the following points. (a) at r = 11.5 cm b) at r = 19.5 cm The important lesson in Gauss^' law is thatthe flux of electric field through a closed surface is set by thenet charge enclosed by the surface. (a) You want the field at agiven radius, which is between the shells. Do you see that you needto use a Gaussian sphere of this radius? How much charge isenclosed by this Gaussian sphere? (b) Now you want the fieldoutside both shells. What Gaussian surface should you now use, andhow much charge does it enclose?

Explanation / Answer

As mentioned in the question , we can use Gauss law to findelectric field . From Gauss Law , net flux through any closed body is equal toQ/ where Q is the net charge enclosed , A)For Electric field at r=11.5 ,    we can consider a spherical concentric sphere ofr=11.5 cm Then Net charge enclosed = 1.50*10-8 C Also as the electric field will be radial and hence will be in linewith the area vector , So Net flux = Electric field * area *cos 0o=E*4r2 From Gauss law , Net flux = 1.50*10-8/ Equating , E*4r2 =1.50*10-8/ E= 1.50*10-8 / 4(11.5*10-2)2 =1.02*104 N/C B)Similarly ,at r=19.5 cm we can consider a spherical concentric sphere of r=19.5 cm Then Net charge enclosed = 1.50*10-8C+2.40*10-8 C=3.90*10-8 C Also as the electric field will be radial and hence will be in linewith the area vector , So Net flux = Electric field * area *cos 0o=E*4r2 From Gauss law , Net flux = 3.90*10-8/ Equating , E*4r2 =3.90*10-8/ E= 3.90*10-8 / 4(19.5*10-2)2 =9.23*103 N/C

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