1.

Suppose [epsilon_(0)] is permittivity of free sapce. If M = mass, L = length, T = time and A = electric current, then.........

Answer»

`|epsilon_(0)| = [M^(-1)L^(-3)T^(2)A]`
`|epsilon_(0)| = [M^(-1)L^(-3)T^(4)A^(2)]`
`|epsilon_(0)| = [M^(-1)L^(-3)T^(4)A^(2)]`
`|epsilon_(0)|= [M^(-1)L^(2)T^(-1)A^(-2)]`

Solution :`F = (q_(1)q_(2))/(4piepsilon_(0)r^(2)) rArr epsilon_(0) = (q_(1)q_(2))/(4piFr^(2))`
`THEREFORE |epsilon_(0)| =(|q_(1)||q_(2)|)/(4piFr^(2))` `[therefore 4PI "dimensionless"]`
`=((AT)(AT))/((M^(1)L^(1)T^(-2))(L^(2))) = M^(-1)L^(-3)T^(4)A^(2)`


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