Community Answer

【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2Given data are\( \begin{align*} \mathrm{{Q}} &= \mathrm{{20}\mu{C}} \end{align*} \)Inner radius \( \begin{align*} \mathrm{{a}} &= \mathrm{{2.5}{c}{m}} \end{align*} \)Outer radius\( \begin{align*} \mathrm{{b}} &= \mathrm{{8}{c}{m}} \end{align*} \)Total charge in the volume is calculated as{:[a=2.5cm],[b=8cm],[Q=20 muC]:}a) r=5cmb r=12cm{:[" charge density "=(" charge ")/(" total vilures ")=(Q)/((4)/(3)pi(b^(3)-a^(3)))=(20 xx10^(-6))/((4)/(3)pi(8^(3)-2.5^(3))xx(10^(-2))],[rho=(3xx20 xx10^(-6))/(4xx3.14 xx(512-15.625)xx10^(-6))=9.624 xx10^(-3)C]:}Using Gam's law ointEds=(1)/(epsi_(0))int rho*dt.CS Scanned with CamScannerExplanation:Please refer to solution in this step.Step2/2The field at any point can be calculated by Gausses law\( \ ... See the full answer