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(D) Given,the volume charge density of the sphere is $ho_v = 1 	imes 10^{-6} \ C/m^3$.The constant $\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9 \ Nm

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$12 \pi$

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(D) Given,the volume charge density of the sphere is $ho_v = 1 	imes 10^{-6} \ C/m^3$.The constant $\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9 \ Nm^2 C^{-2}$.The distance from the centre is $r = 1 \ mm = 10^{-3} \ m$.For a point inside a uniformly charged sphere,the electric field $E$ is given by Gauss's Law as:$E = \frac{ho_v r}{3 \epsilon_0}$Since $\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9$,we have $\frac{1}{\epsilon_0} = 36 \pi 	imes 10^9$.Substituting this into the formula:$E = \frac{ho_v r}{3} 	imes (36 \pi 	imes 10^9)$$E = ho_v r 	imes 12 \pi 	imes 10^9$Substituting the values of $ho_v$ and $r$:$E = (1 	imes 10^{-6}) 	imes (10^{-3}) 	imes 12 \pi 	imes 10^9$$E = 10^{-9} 	imes 12 \pi 	imes 10^9$$E = 12 \pi \ N/C$.Therefore,the electric field is $12 \pi \ N/C$.

$A$ spherical volume contains a uniformly distributed charge of density $1.0 \times 10^{-6} \ C/m^3$. Find the electric field (in $N/C$) at a point inside the volume at a distance $1 \ mm$ from the centre. (Let $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \ Nm^2 C^{-2}$)

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