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Question
A cube of marble having each side 1 cm is kept in an electric field of intensity 300 V/m. Determine the energy contained in the cube of dielectric constant 8. [Given : ∈0 = 8.85 10–12 C^2/Nm^2]
Solution
Given: `l = 1 cm = 10^-2 m, E = 300 Vm^-11, k = 8`
To find: Energy contained in the cube (U)
formula: `u=U/V`
volume of marble
`V=(i^3)=(10^-2)3=10^-6m^3`
Energy Density, `u=1/2 in_0kE^2`
`=1/2xx8.85xx10^-12xx8xx(300)^2`
`=3.185xx10^-6 J/m^3`
From formula `U=u xx V=3.185xx10^-6xx10^-6`
`U=3.185xx10^-12 J`
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