, . ( ) 1) Original question Problem 1 (Gauss & Graphs). At r=a\approx 3.0\ \text{cm} the graph shows the inside field E_{\text{in}}(a) and the outside field E_{\text{out}}(a). Vertical scale unit is E_s=1.00\times10^{8}\ \text{N/C}. 2) , E-r ( E_s) r=a / . 3) E(r)=\dfrac{kQ_{\text{enc}}}{r^2} ( k=\dfrac1{4\pi\varepsilon_0} ). ( ra ): Q_{\text{enc}}=q+Q_{\text{shell}}\Rightarrow E_{\text{out}}(a)=\dfrac{k(q+Q_{\text{shell}})}{a^2}. q=\frac{E_{\text{in}}(a)a^2}{k},\qquad Q_{\text{shell}}=\frac{\big(E_{\text{out}}(a)-E_{\text{in}}(a)\big)a^2}{k}. 4) ( ) r=a 3.0\ \text{cm}. E_{\text{in}}(a)\approx 0.60\,E_s, E_{\text{out}}(a)\approx 0.25\,E_s. ( 0.02 ) / a=3.0\ \text{cm}=0.030\ \text{m} E_s=1.00\times10^{8}\ \text{N/C} k=8.9876\times10^{9}\ \text{Nm}^2/\text{C}^2 (A) q \begin{aligned} q &=\frac{E_{\text{in}}(a)a^2}{k} =\frac{(0.60)(1.00\times10^{8})(0.030)^2}{8.9876\times10^{9}}\\ &=6.01\times10^{-6}\ \text{C} =6.0\ \mu\text{C}\ (\text{}) \end{aligned} (B) Q_{\text{shell}} \begin{aligned} Q_{\text{shell}} &=\frac{(E_{\text{out}}-E_{\text{in}})a^2}{k} =\frac{(0.25-0.60)(1.00\times10^{8})(0.030)^2}{8.9876\times10^{9}}\\ &=-3.50\times10^{-6}\ \text{C} =-3.5\ \mu\text{C}. \end{aligned} (A) q \approx +6.0\ \mu\text{C} (B) Q_{\text{shell}} \approx -3.5\ \mu\text{C} : E_{\t