A hollow Spherical conductor of radius 12 cm is

Physics

1 answer:

Ket [755]2 years ago

5 0

Answer:

Explanation:

Radius of the charged sphere (R)=12cm

Charge (q)=6∗10^−6C

Electric field intensity (E)=?

i) On the surface

$E=\frac{1}{4\pix_{e0} } \frac{q}{R^2} =9*10^{9} *\frac{6*10^{-6} }{0.12^-2} =3.75*10^6 N/C$

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

$E=\frac{1}{4\pi _{e} }_{0} \frac{q}{r^{2} } =9*10^9*\frac{6 \a* *10^6}{(0.27)^2} =7.41*10^5 N/C$

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