Answer:
The electric field is [tex]E = 2.2625 *10^{6} \ N/C[/tex]
Explanation:
From the question we are told that
The radius of the inner sphere is [tex]r_1 = 0.008\ m[/tex]
The radius of the outer sphere is [tex]r _2 = 0.018 \ m[/tex]
The charge on the inner sphere is [tex]q_1 = 3.62 *10^{-8} \ C[/tex]
The charge on the outer sphere is [tex]q_2 = 1.62 *10^{-8} \ C[/tex]
The position from the origin is [tex]d = 0.012 \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{k (q_1 )}{ r^2}[/tex]
The reason for using [tex]q_1[/tex] for the calculation is due to the fact that the position considered is greater than the [tex]r_1[/tex] but less than [tex]r_2[/tex]
Here k is the Coulomb constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A{-2}[/tex]
So
[tex]E = \frac{9*10^9 (3.62 *10^{-8}}{0.012^2}[/tex]
[tex]E = 2.2625 *10^{6} \ N/C[/tex]