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 : `epsilon_(0) =8.85 xx 10^(-12) C^(2)//Nm^(2)`]

Answer 1

Given, `k=8`
`E=300V//m`
`l=1cm=10^(-2)m`
Volume of the cube,
`V=(1cm)^(3)=(10^(-2)m)^(3)=10^(-6)m^(3)`
`epsilon_(0)=8.85 xx 10^(-12)C^(2)//Nm^(2)`
Energy stored per unit volume,
`U=(1)/(2)k epsilon_(0)E^(2)`
`U=(1)/(2) xx 8xx8.85 xx 10^(-12)xx(300)^(2)`
`U=3.18xx10^(-8)J//m^(3)`
`therefore` Energy contained in cube
`=U xx V`
`=318.6xx10^(-8)xx10^(-6)`
`=318.6xx10^(-14)J//m^(3)`