Answer:
a)E= 0
b) [tex]E=\dfrac{Q}{\varepsilon _o\times 4\pi a^2}\ N/C[/tex]
Explanation:
Given that
Charge Q is distributed on a metallic sphere of radius a
a)r < a.
At a radius r ,from gauss theorem
[tex]E.ds=\dfrac{q_i}{\varepsilon _o}[/tex]
But in the sphere there is no any charge inside the sphere so
[tex]E.ds=\dfrac{o}{\varepsilon _o}[/tex]
E.ds = 0
E= 0
b) r > a
At a radius r ,from gauss theorem
[tex]E.ds=\dfrac{q_i}{\varepsilon _o}[/tex]
[tex]E\times 4\pi a^2=\dfrac{Q}{\varepsilon _o}[/tex]
[tex]E=\dfrac{Q}{\varepsilon _o\times 4\pi a^2}\ N/C[/tex]
