The electric force between two electric charges is given by `F= (1)/(4pi epsi_(0))(q_(1)q_(2))/(r^(2))`. where is a distance . between charges `q_(1) and q_(2)` . So, unit and dimensional formula of `epsi_(0)`, is ...... and...... respectively.

Options

A. `N^(1)C^(2)m^(-2),M^(-1)L^(-3)T^(4)A^(2)`
B. `N^(-1)Cm^(-2),M^(-1)L^(-1)T^(-1)A`
C. `N^(2)C^(2)m^(-2)M^(-1)L^(-3)T^(4)A^(2)`
D. `NCm^(-2),MLTA`

Explanation

`F=(1)/(4pi epsi_(0)).(q_(1)q_(2))/(r^(2))`
`:.epsi_(0)=(q_(1)q_(2))/(4pi Fr^(2))`
`:.` Units of `epsi_(0)=(C^(2))/(Nm^(2))=N^(-1)C^(2)m^(-2)`
`[epsi_(0)]=([C^(2)])/([N][m^(2)])=(A^(2)T^(2))/((M^(1)L^(1)T^(-2))(L^(2)))`
`=M^(-1)L^(-3)T^(4)A^(2)`