The dimensions of `1/2 epsilon_(0)E^(2) (epsilon_(0)=` permittivity of free space, `E=` electric field) is

Options

A. `ML^(-1)T^(-2)`
B. `ML^(-1)T^(-1)`
C. `MLT^(-2)`
D. None of these

Answer

A

Explanation

`[u]=[(1)/(2)varepsilon_(0)E^(2)]`
`implies [u]=[M^(-1)L^(-3)A^(2)T^(4)][MLT^(-3)A^(-1)]^(2)`
`[u]= [ML^(-1)T^(-2)]`
Hence option a is correct. In class XII we shall study that given expression stands for energy per unit volume and so we can evaluate dimensions on that basis also.