In the figure below, a solid sphere, of radius a = 2.60 cm is concentric with a
ID: 2132724 • Letter: I
Question
In the figure below, a solid sphere, of radius a = 2.60 cm is concentric with a spherical conducting shell of inner radius b = 2.00a and outer radius c = 2.40a. The sphere has a net charge q1 = +5.00 fC which is distributed uniformly through the sphere; the shell has a net charge of q2 = -q1.
Explanation / Answer
radius of the sold sphere a =2.6 cm
inner radius of spherical conducting shell b=2a
outer radius of spherical conducting shell c=2.4a
the net charge of sphere q1 =+5 fC =5*10^-15 C
the net charge of conducting shell q2 =-q1
apply Gauss' law,
?E.ds =q/?_0 .......... (1)
?E.ds =(4?r^2)E ......... (2) (since ? ds =4?r^2)
compare equation (1) and (2),we get
(4?r^2)E = q/?_0
therefore, electric field E =(1/4??_0)(q/r^2) ......... (3)
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a)
at r =0 cm and a =2.6 cm
at r=0 the charge q =0
therefore electric field E =0
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b)
given r =a/4 ........ (4)
if r<a then electric field E =(1/4??_0)(q'/r^2) ....... (5)
where, charge q' =q(r/a)^3 ...... (6)
eq (5), becomes
E =(1/4??_0)(qr^3/a^3r^2)
E =(1/4??_0)(qr/a^3) .......... (7)
eq (7), becomes
E =(1/4??_0)(q/3a^2) ........ (8)
where, 1/4??_0 =8.99*10^9 N.m^2/C^2
substitute the given data in eq (8), we get
E =2.22*10-2 N/C
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c)
given r= a =26 cm =2.6*10^-2 m
electric field E =kq/a^2 ......... (9)
E =6.66*10^-2 N/C
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d)
given r =1.2 a =1.2*2.6 =3.12*10^-2 m
electric field E = E =kq/r^2 at a<r<b
E =4.62*10^-2 N/C
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e)
given r =2.20a =5.72*10^-2 m
if b<r<c, then the electric field E =0
since chrge q does not lie inside the shell (q =0).
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f)
given r =3a =7.8*10^-2 m
the charge q1 =+5 fC lies inside the shell and q2 =-5 fC lies on the out side
so the charge enclosed by the Gaussian surface is zero.
therefore electric field E =0.
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g)
the net charge on the inner surface is q2+q1 =0
q2=-q1 =-5 fC
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h)the net charge on the outer surface is
q_0 =-q2-q1
q_0 =q1-q1 (since q2 =-q1)
q_0 =0 fC