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: $ϵ_{0} = 8.85 \times 10^{- 12} C^{2} / N m^{2}$ ].

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Solution

Electric field $E = 300$ V/m
Length of the cube $l = 0.01 m$
Thus volume of cube $V = l^{3} = 10^{- 6} m^{3}$
Dielectric constant $K = 8$
Energy contained in the cube $U = \frac{1}{2} ϵ E^{2} V$
Or $U = \frac{1}{2} K ϵ_{o} E^{2} V$
$∴$ $U = \frac{1}{2} \times 8 \times 8.85 \times 10^{- 12} \times (300)^{2} \times 10^{- 6}$
$⟹$ $E = 3.186 \times 10^{- 12}$ $J$

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