Explanation:
The given data is as follows.
q = 6.0 nC = [tex]6 \times 10^{-9} C[/tex]
inner radius (r) = 1.0 cm = 0.01 m (as 1 cm = 100 m)
So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.
Formula to calculate the charge density is as follows.
[tex]\sigma = \frac{q_{in}}{A}[/tex] .......... (1)
Since, area of the sphere is as follows.
A = [tex]4 \pi r^{2}[/tex] ........... (2)
Hence, substituting equation (2) in equation (1) as follows.
[tex]\sigma = \frac{q_{in}}{4 \pi r^{2}}[/tex]
= [tex]\frac{6 \times 10^{-9} C}{4 \times 3.14 \times (0.01)^{2}}[/tex]
= [tex]0.477 \times 10^{-5}[/tex]
or, = 4.77 [tex]\mu C/m^{2}[/tex]
Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 [tex]\mu C/m^{2}[/tex].
