Answer:
Explanation:
Given
radius [tex]r=0.227 m[/tex]
Charge on surface [tex]Q=6.03\times 10^{-6} C[/tex]
Point Charge inside sphere [tex]q=1.15\times 10^{-6} C[/tex]
Electric Field at [tex]r=0.735 m[/tex]
Treating Surface charge as Point charge and applying Gauss law
[tex]E_{total}A=\frac{q_{enclosed}}{\epsilon _0}[/tex]
where A=surface area up to distance r
[tex]E_{total}=\frac{Q+q}{4\pi r^2}[/tex]
[tex]E_{total}=\frac{6.03\times 10^{-6}+1.15\times 10^{-6}}{4\pi (0.735)^2\times 8.85\times 10^{-12}}[/tex]
[tex]E_{total}=1.194\times 10^{5} N/C[/tex]