What is the product of permeability of free space and permittivity?
Answers
Answer:
hope this will help you.
We know that
k = coulomb constant
= 1 / 4 × Pie × € = 9 × 10^9
=> 1 / 4 × Pie × € = 9 × 10^9
Taking reciprocal on both sides
=> 4 × Pie × € = 1 / ( 9 × 10^9 )
=> € = 1 / ( 4 × Pie × 9 × 10^9 ) -- (i)
where € = permittivity of free space
also u = 4 × Pie × 10^ -7 ---(ii)
Where u = permeability of free space
On multiplying equation (i) and (ii) gives
€ × u
= [ 1 / ( 4 × Pie × 9 × 10^9 ) ] × [ 4 × Pie × 10^ -7]
= ( 4 × Pie × 10^ -7 ) / (4 × Pie × 9 × 10^9 )
[ Note : 4 × Pie gets cancelled from Numerator and denominator ]
= ( 9 × 10^9 ) / ( 10^ -7 )
= 9 × 10 ^16
=> € × u = 9 × 10 ^16
= [ 3 × 10^8 ] ^2 = c ^2
where c = speed of light
Thus , product of permitivity and permeability is equal to the square of the speed of light...
Taking square Root on both sides
=> Root ( € × u ) = Root ( 9 × 10 ^16 )
=> Root ( € × u ) = 3 × 10^8 = c
where c = Speed of light