Question

Calculate the energy stored in the capacitor per unit length, in units of $J/m$.

Answer 1

Capacitance C =$\frac{2\pi {\epsilon }_0}{\ln \frac{R_0}{R_1}}$

=$\frac{1}{2\times 9\times {10}^9\times \ln \frac{75}{18}}=0.04\times {10}^{-9}=40$RF

Energy stored E = $\frac{Q^2}{2C}=\frac{{30}^2}{2\times 40\times {10}^{-}12}=11.25\times {10}^{12}$ J