In the figure below, a solid sphere, of radius a = 1.90 cm is concentric with a
ID: 2003119 • Letter: I
Question
In the figure below, a solid sphere, of radius a = 1.90 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 chargeq1 = +6.00 fC which is distributed uniformly through the sphere; the shell has a net charge of q2 = -q1.
(a) What is the magnitude of the electric field at radial distance r = 0?
= _____ N/C
(b) What is the magnitude of the electric field at radial distance r = a/3?
= _____ N/C
(c) What is the magnitude of the electric field at radial distance r = a?
= _____ N/C
(d) What is the magnitude of the electric field at radial distance r = 1.30a?
= _____ N/C
(e) What is the magnitude of the electric field at radial distance r = 2.10a?
= _____ N/C
(f) What is the magnitude of the electric field at radial distance r = 3.90a?
= _____ N/C
(g) What is the net charge on the inner surface of the shell?
= _____ fC
(h) What is the net charge on the outer surface of the shell?
= _____ fC
Explanation / Answer
Radius of the sold sphere a =1.9 cm
Inner radius of spherical conducting shell b=2a
Outer radius of spherical conducting shell c=2.4a
The net charge of sphere q1 =+6 fC =6*10^-15 C
The net charge of conducting shell q2 =-q1
Applying Gauss' law,
E.ds =q/_0 .......... (1)
E.ds =(4r^2)E ......... (2) (since ds =4r^2)
compare equation (1) and (2),we get
(4r^2)E = q/_0
therefore, electric field E =(1/4_0)(q/r^2) ......... (3)
a)
At r =0 cm and a =1.9 cm
At r=0 the charge q =0
Therefore electric field E =0
b)
Given r =a/3 ........ (4)
It means r<3
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 =4.962*10-2 N/C
c)
Given r= a =1.9 cm =1.9*10^-2 m
Electric field E =kq/a^2 ......... (9)
E =14.93*10^-2 N/C
d)
Given r =1.3 a =1.3*1.9 =2.47*10^-2 m
Electric field E = E =kq/r^2 at a<r<b
E =8.85*10^-2 N/C
e)
Given r =2.10a =3.99*10^-2 m
If b<r<c, then the electric field E =0
Since chrge q does not lie inside the shell (q =0).
f)
Given r =3.4a =6.46*10^-2 m
The charge q1 =+6 fC lies inside the shell and q2 =-6 fC lies on the out side
So the charge enclosed by the Gaussian surface is zero.
Therefore electric field E =0.
g)
The net charge on the inner surface is q2+q1 =0
q2=-q1 =-6 fC
The net charge on the outer surface is
q_0 =-q2-q1
q_0 =q1-q1 (since q2 =-q1)
q_0 =0 fC