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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. - Physics

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]

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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`

 

Concept: Energy of Charged Condenser
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