Physics - Electrostatics Question with Solution | TestHub

Energy of a charged capacitor,

$E=\frac{1}{2}\frac{Q^2}{C}$

For a cylindrical capacitor,
$C= \frac{2\pi {\epsilon }_{0}L}{{\log }_{\mathrm{e}}\left(\frac{b}{a}\right)}$

$E^{}= \frac{1}{2}\frac{Q^2}{2\pi {\epsilon }_{0}L}{\log }_e\left(\frac{b}{a}\right)$...(i)

where$L$= length of the cylinder

$a$and$b$= radii of two concentric cylinder

$C^{\text{'}}= \frac{2\pi {\epsilon }_0\left(2L\right)}{{\log }_e\left(\frac{b}{a}\right)}$

$E^{\text{'}}=\frac{1}{2}\frac{{\left(2Q\right)}^2}{C\text{'}}$

$=\frac{1}{2}\frac{{\left(2Q\right)}^2}{2\pi {\epsilon }_0\left(2L\right)}{\log }_e\left(\frac{b}{a}\right)$...(ii)

From Eqs. (i) and (ii), we get

$E^{\text{'}}= 2E$