$A$ point charge causes an electric flux of $-1.0 \times 10^{3} \; N \cdot m^{2} / C$ to pass through a spherical Gaussian surface of $10.0 \; cm$ radius centered on the charge.
$(a)$ If the radius of the Gaussian surface were doubled,how much flux would pass through the surface?
$(b)$ What is the value of the point charge?

(B) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_{0}}$. This flux depends only on the net charge enclosed by the surface and is independent of the size or shape of the Gaussian surface. Therefore,if the radius is doubled,the flux remains unchanged at $-1.0 \times 10^{3} \; N \cdot m^{2} / C$.
$(b)$ Using the relation $\phi = \frac{q}{\varepsilon_{0}}$,we can find the charge $q$ as $q = \phi \varepsilon_{0}$.
Given $\phi = -1.0 \times 10^{3} \; N \cdot m^{2} / C$ and $\varepsilon_{0} = 8.854 \times 10^{-12} \; C^{2} / (N \cdot m^{2})$:
$q = (-1.0 \times 10^{3}) \times (8.854 \times 10^{-12}) = -8.854 \times 10^{-9} \; C = -8.854 \; nC$.
Thus,the value of the point charge is $-8.854 \; nC$.