Answer :
Answer:
E = 53.3 N/C
Explanation:
The charge of the sphere can be found by Gauss’ Law:
[tex]\int\vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}[/tex]
We will draw an imaginary spherical surface around the original sphere with radius 0.2 cm. The integral in the left-hand side is the area of this surface, therefore no need to take the integral. The right-hand side is equal to the total charge enclosed by this imaginary surface.
[tex]E2\pi r^2 = \frac{Q_{enc}}{\epsilon_0}\\(480)4\pi(0.002)^2 = \frac{Q}{8.8\times 10^{-12}}\\Q = 2.12\times 10^{-13}[/tex]
Now that we have found the charge of the sphere, we can now apply Gauss’ Law again to find the E-field 0.006 m from the center.
We will draw our second imaginary spherical surface with a radius 0.006 m.
[tex]E4\pi r^2 = \frac{Q}{\epsilon_0}\\E4\pi(0.006)^2 = \frac{2.12 \times 10^{-13}}{\epsilon_0}\\E = 53.3~{\rm N/C}[/tex]