Answer
(a) $ \Phi_{E} = 3.53 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$
(b) $Q = 3.13 \times 10^{-6} \mathrm{C}$
Work Step by Step
(a) The electric flux through the sphere will be \begin{aligned} \Phi_{E} &=E A=E\left(4 \pi r^{2}\right) \\ &=\left(1.25 \times 10^{6} \mathrm{N} / \mathrm{C}\right)4 \pi(0.15 \mathrm{m})^{2} \\ &=3.53 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} \end{aligned} (b) The electric field is related to the charge $Q$ by \begin{aligned} Q &=\frac{r^{2} E}{k} \\ &=\frac{(0.15 \mathrm{m})^{2}\left(1.25 \times 10^{6} \mathrm{N} / \mathrm{C}\right)}{\left(9.0 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)} \\ &=3.13 \times 10^{-6} \mathrm{C} \end{aligned}
