The energy density at a point in a medium of dielectric constant 6 is `26.55xx10^(6)J//m^(3)`. What is the electric field intensity at that point ? `[epsi_(0)=8.85xx10^(-12)SI` units]

Options

A. `2xx10^(8)N//C`
B. `10^(9)N//C`
C. `3xx10^(9)N//C`
D. `sqrt(10^(9)N//C)`

Answer

B

Explanation

Energy density `=(1)/(2)epsi_(0)KE^(2)`
`therefore E=sqrt((2("energy density"))/(epsi_(0)K))`
`=sqrt((2xx26.55xx10^(6))/(8.85xx10^(-12)xx6))`
`E=sqrt(10^(12+16))=10^(9)N//C`