Question

In the figure below, a solid sphere, of radius a = 2.90 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 charge q1 = +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?
1 N/C

(b) What is the magnitude of the electric field at radial distance r = a/3?
2 N/C

(c) What is the magnitude of the electric field at radial distance r = a?
3 N/C

(d) What is the magnitude of the electric field at radial distance r = 1.30a?
4 N/C

(e) What is the magnitude of the electric field at radial distance r = 2.10a?
5 N/C

(f) What is the magnitude of the electric field at radial distance r = 3.40a?
6 N/C

(g) What is the net charge on the inner surface of the shell?
7 fC

(h) What is the net charge on the outer surface of the shell?
8 fC

Answer 1

radius of the sold sphere a =2.9 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

apply Gauss' law,

∫E.ds =q/ε_0 .......... (1)

∫E.ds =(4πr^2)E    ......... (2)   (since ∫ ds =4πr^2)  

compare equation (1) and (2),we get

(4πr^2)E = q/ε_0

therefore, electric field E =(1/4πε_0)(q/r^2)    ......... (3)

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a)

at r =0 cm and a =2.9 cm

at r=0 the charge q =0

therefore electric field E =0

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b)

given r =a/3   ........ (4)

its means r<3

if r

where, charge q' =q(r/a)^3 ...... (6)

eq (5), becomes

E =(1/4πε_0)(qr^3/a^3r^2)

E =(1/4πε_0)(qr/a^3) .......... (7)

eq (7), becomes

E =(1/4πε_0)(q/3a^2) ........ (8)

where, 1/4πε_0 =8.99*10^9 N.m^2/C^2

substitute the given data in eq (8), we get

E =2.13*10-2 N/C

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c)

given r= a =2.9 cm =2.9*10^-2 m

electric field E =kq/a^2 ......... (9)

E =6.41*10^-2 N/C

.........................................................

d)

given r =1.3 a =1.3*2.9 =3.77*10^-2 m

electric field E = E =kq/r^2    at a

E =3.8*10^-2 N/C

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e)

given r =2.10a =6.09*10^-2 m  

if b

since chrge q does not lie inside the shell (q =0).

............................................................

f)

given r =3.4a =9.86*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.

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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