Question

A cube of marble having each side $1$cm is kepy in an electric field of intensity $300$ V$/$m. Determine the energy contained in the cube of dielectric constant $8$. [Given: ${ϵ}_0=8.85\times {10}^{-12}{C}^2/N{m}^2$].

Answer 1

Electric field $E=300$ V/m

Length of the cube $l=0.01m$

Thus volume of cube $V=l^3={10}^{-6}{m}^3$

Dielectric constant $K=8$

Energy contained in the cube $U=\frac{1}{2}ϵE^{2}V$

Or $U=\frac{1}{2}K{ϵ}_o{E}^{2}V$

$\therefore$ $U=\frac{1}{2}\times 8\times 8.85\times {10}^{-12}\times (300{)}^2\times {10}^{-6}$

$⟹$ $E=3.186\times {10}^{-12}$ $J$