Question

Calculate the energy stored per unit length in the capacitor in units of joules per meter.

Answer 1

Given :

$\lambda =30$ C/m

$R_1=18$ cm

$R_2=75$ cm

Capacitance of cylindrical capacitors

$C=\frac{2\pi {ϵ}_{o}L}{\ln \frac{R_2}{R_1}}$

where $L$ is the length of the capacitor

Energy stored

$E=\frac{Q^2}{2C}=\frac{(\lambda L{)}^2\times \ln \frac{R_2}{R_1}}{2\pi {ϵ}_{o}L}$

$⟹$ Energy stored per unit length

$\frac{E}{L}=\frac{{\lambda }^2\times \ln \frac{R_2}{R_1}}{2\pi {ϵ}_o}=\frac{(30{)}^2\times \ln \frac{75}{18}}{2\pi (8.85\times {10}^{-12})}=2.31\times {10}^{13}$ $\frac{J}{m}$