Answer:
a
The value at an inside point is Zero
b
The electric field is [tex]E = 2.7*10^{6} \ N/C[/tex]
Explanation:
From the question we are told that
The magnitude of the charge is [tex]q = 3.0 \mu C = 3.0 *10^{-6} \ C[/tex]
The radius of the spherical ball is [tex]r = 5.0 \ mm = 0.005 \ m[/tex]
Generally according to law postulated by Gauss the magnitude of charge enclosed inside a conducting material is zero which implies that the electric field inside the spherical ball is zero
Generally the electric field out side the spherical ball is mathematically represented as
[tex]E = \frac{kq}{ a^2}[/tex]
Here a is the position outside the spherical ball that is been considered and the value is [tex]a = 10 \ cm = \frac{10}{100} = 0.1 \ m[/tex]
and k is the coulombs constant with value
[tex]k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
=> [tex]E = \frac{ 3 *10^{-6} * 9*10^9 }{ (0.1)^2}[/tex]
=> [tex]E = 2.7*10^{6} \ N/C[/tex]
