Respuesta :
Answer: 446.28*10^3 N/C ( directed to center of the sphere)
Explanation: In order to explain this problem we have to use the electric field expression, which is given by:
E=k*q/r^2 where k is a constant equal 9*10^9 N/C^2*m^2, r is the distance from the charge to calculate the electric field.
Then we have,
E=(9*10^9*0.6*10^-6)/(0.11)^2= 446.28*10^3 N/C
Due to the negative point charge, the electric field is directed to the center of the sphere.
The electric field at a distance of 0.11 m from the center of the sphere is [tex]446.28\times 10^3\;\rm N/C[/tex].
How do you calculate the electric field from the center of the sphere?
Given that the inner radius r1 of the sphere is 0.200 m and the outer radius r2 of the sphere is 0.240 m. The charge density at the inner surface is +6.67 × 10^−6 C/m^2. The charge at the outer surface is -0.600 μC.
The charge density at the outer surface is calculated as given below.
[tex]\sigma _2 = \dfrac {q}{4\pi r_2^2}[/tex]
[tex]\sigma _2 = \dfrac {-0.6\times 10^{-6}}{4\times 3.14\times 0.240^2}[/tex]
[tex]\sigma_2 = -8.29 \times 10^{-7}\;\rm C/m^2[/tex]
The outer surface has a negative point charge then the electric field is directed to the center of the sphere which is calculated as given below.
[tex]E =k \dfrac {q}{r^2}[/tex]
Where q is the charge at the outer surface, r is the distance from the center and k is constant which is equal to 9*10^9 N/C^2*m^2.
[tex]E = 9\times 10^9\times \dfrac {0.6\times 10^{-6}}{0.11^2}[/tex]
[tex]E = 446.28 \times 10^3\;\rm N/C[/tex]
Hence we can conclude that the electric field at a distance of 0.11 m from the center of the sphere is [tex]446.28\times 10^3\;\rm N/C[/tex].
To know more about the electric field, follow the link given below.
https://brainly.com/question/12757739.
