Question

Consider a solid, insulating sphere of radius a = 5.00 cm carrying a net positive charge of Q = 3.00 μC uniformly distributed throughout its volume. Concentric with this sphere is a hollow, conducting spherical shell with inner radius b = 10.0 cm and outer radius c = 15.0 cm having a net negative charge of q = -1.00 μC.

a) Give the expression from Gauss’s Law for the magnitude of electric field inside the insulating sphere for 0 ≤ r ≤ a, and use this expression to calculate the value of the electric field at r = 0 and r = a = 5.00 cm.

b) Since the spherical shell is conducting, it contributes nothing to the electric field inside its hollow space. Give the expression from Gauss’s Law for the magnitude of electric field for a ≤ r < b, and use this expression to calculate the value of the electric field at r = a = 5.00 cm and r = b = 10.0 cm. (Note your value here of the electric field at r = a should agree with your result above.)

c) From the properties of conductors in electrostatic equilibrium, what is the magnitude of the electric field inside the conducting shell for b < r < c?

d) Because of your answer above, what must be the charge on the inner surface of the hollow conducting shell?

e) Given the net charge carried by the hollow conducting shell, what must be the charge on its outer surface?

f) Using Gauss’s Law, determine the expression for the magnitude of electric field for r > c, and use this expression to calculate the value of the electric field at r = c = 15.00 cm and r = 25.0 cm.

g) Make a graph of the magnitude of the electric field for 0 ≤ r ≤ 25.0 cm. Draw this to scale specifying your results calculated above.

Answer 1

Figure shows system of charged solid sphere (insulator) of radius 5 cm surrounded by a

charged concentric spherical shell (conductor ) of inner radius 10 cm and outer radius 15 cm.

It is given that solid sphere is holding charge Q = 3 C and the charge is distributed

uniformly throughout the volume. Volume density of charge = { 3 10^-6 / [ 4/3 125 ] } C/ cm³

we get = 5.7 10^-3 C / m³

To get electric field inside solid sohere, i.e. radial distance r < 5, we imagine a gaussian spherical surface ( not shown in figure) of radius r.

enclosed charge inside the imagined gaussian surface is (4/3)r³ = (3/4) r³ C
Hence electric field intensity E(r) inside solid sphere is obtained from Gauss law as

E(r) 4r² = [3/ 4]r³ /

Hence, E(r) = [ ( 0.24 )/ ] r , r < a

E(r) at r=0   is zero

E(r) at r=5 cm E(5) = ( 0.24 5.7 10^-3 )/ ( 8.854 10^-12 ) = 1.54 10⁸ V/m

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Between the solid sphere and spherical shell , we get electric field intensity by considering a spherical Gaussian surface G₁ as shown in figure. If the radius of G₂ is r so that a < r < b, then from Gauss law, we get electric field intensity E(r) as

E(r) 4r² = Q /     or    E(r) = Q / [ (4 ) r² ]   , a < r < b

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Since spherical shell is conductor, electric field is zero inside spherical shell.

To get zero electric field, the net enclosed charge enclosed inside Gaussian spherical surface G₂ should be zero.

Since the solid sphere is having charge +3 , then the charge induced in the inner surface of spherical shell will be -3 so that net enclosed charge is zero.

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If net charge of spherical shell is -1 and inner surface has induced charge -3 ,

then charge on the outer surface = +2

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Electric field intensity outside spherical shell is obtained by considering a Gaussain spherical surface of radius greater than 15 cm as given below

E(r) 4r² = ( Q + q ) /   

where Q+q is the net charge enclosed by gaussian surface, Q = +3 is the charge contained in solid sphere

and q = -1 is net charge contained in spherical shell.

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Electric field intensity as a function of radial distance is plotted and shown in figure