Answer:
Explanation:
Radius of the charged sphere (R)=12cm
Charge (q)=6∗10^−6C
Electric field intensity (E)=?
i) On the surface
[tex]E=\frac{1}{4\pix_{e0} } \frac{q}{R^2} =9*10^{9} *\frac{6*10^{-6} }{0.12^-2} =3.75*10^6 N/C[/tex]
ii) Inside the sphere,
E=0; Because charge enclosed by Gaussian surface is zero.
iii) Outside the sphere at distance 15cm
i.e. r=12+15=27cm=0.27m
[tex]E=\frac{1}{4\pi _{e} }_{0} \frac{q}{r^{2} } =9*10^9*\frac{6 \a* *10^6}{(0.27)^2} =7.41*10^5 N/C[/tex]
