3) (15 points) The figure below illustrates a small solid conducting sphere havi
ID: 1881045 • Letter: 3
Question
3) (15 points) The figure below illustrates a small solid conducting sphere having radius R, and charge Qi. A larger concentric infinitesimally thin insulating shell has uniformly distributed charge Q2 and radius R2> R. An even larger thick concentric charged conducting shell has inner radius R > R, outer radius R4> R3, and charge Qs. Furthermore, Q QQ3Q, and Q, 4Q, where Q is a positive quantity Use the appropriate laws of physics to determine expressions Q, r, Ri, R2, R3, Ra, and eo) for the following: a) the magnitude of the electric field in all 5 regions (r Ri. b) the net charge on the inner and outer surfaces of the outer conducting shell (provide sufficient explanation).Explanation / Answer
part a:
for r<R1:
total charge enclosed=Q1*(r/R1)^3
let electric field be E.
using Gauss’ law:
epsilon*E*4*pi*r^2=charge enclosed
where epsilon=electrical permitivity of the free space
==>epsilon*E*4*pi*r^2=Q1*r^3/R1^3
==>E=Q1*r/(4*pi*epsilon*R1^3)
for R1<r<R2:
charge enclosed=Q1
using Gauss’ law:
epsilon*E*4*pi*r^2=Q1
==>E=Q/(4*pi*epsilon*r^2)
for R2<r<R3:
total charge enclosed=Q1+Q2=-2*Q
using Gauss’ law:
epsilon*E*4*pi*r^2=-2*Q
==>E=-Q/(2*pi*epsilon*r^2)
for R3<r<R4:
total charge enclosed=Q4=Q1+Q2+Q3*((r^2-R3^2)/(R4^2-R3^2))
Q4=-2*Q+4*Q*((r^2-R3^2)/(R4^2-R3^2))
using Gauss’ law:
epsilon*E*4*pi*r^2=total charge enclosed=Q4
==>E=Q4/(4*pi*epsilon*r^2)
for r>R4:
total charge enclosed=Q1+Q2+Q3=2*Q
using Gauss' law:
epsilon*E*4*pi*r^2=charge enclosed=2*Q
==>E=Q/(2*pi*epsilon*r^2)
part b:
charge on inner surface on the shell of radius R2=-Q1=-Q
charge on outer surface on the shell of radius R2=Q1+Q2=-2*Q
charge on lower surface on the shell of radius R3=-(-2*Q)=2*Q
charge on the outer surface on the shell of radius R3=-(2*Q)
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