Answer:
1.23 × 10⁷ N/C
Explanation:
Given that:
The radius of the solid sphere R₁ = 2.00 cm
The charge q₁ = 8.95 × 10⁻⁶ C
The radius of the inner spherical shell R₂ = 4.00 cm
The radius of the outer spherical shell R₂ = 5.00 cm
The charge q₂ = -2.33 × 10⁻⁶ C
Assuming the radius of the configuration = 7 cm
Then, the net charge of the system is
[tex]Q_{net} = q_1 +q_2[/tex]
[tex]Q_{net} =[/tex] 8.95 + (-2.33)
[tex]Q_{net} =[/tex] 6.62 µC
Thus, the electric field for the system can be calculated by using the formula;
[tex]E = k \dfrac{Q_{net}}{R^2}[/tex]
[tex]E = \dfrac{(9\times 10^9) \times (6.62 \times 10^{-6})}{(7\times 10^{-2})^2}[/tex]
[tex]E = 12159183.67 \ N/C[/tex]
E = 1.23 × 10⁷ N/C
