Answer:
a)[tex]E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C[/tex]
b)E=0
c)[tex]E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C[/tex]
Explanation:
Given that
A point charge Q is placed at the center of a conducting spherical shell .Due to this - Q charge will induce on the inner sphere surface and +Q will induce on the outer sphere surface .
a) r < a
At a radius r ,from gauss theorem
[tex]E.ds=\dfrac{q_i}{\varepsilon _o}[/tex]
[tex]E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}[/tex]
[tex]E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C[/tex]
b) a < r < b
[tex]E.ds=\dfrac{q_i}{\varepsilon _o}[/tex]
The total induce in this surface = - Q+ Q =0
[tex]E.ds=\dfrac{0}{\varepsilon _o}[/tex]
E = 0
c) r > b
[tex]E.ds=\dfrac{q_i}{\varepsilon _o}[/tex]
[tex]E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}[/tex]
[tex]E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C[/tex]
