Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimension of permittivity is

Question

Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimension of permittivity is (A) ML -2 T 2 A (B) M -1 L -3 T 4 A 2 (C) MLT -2 A (D) ML2 T-1 A2

Tardigrade Admin · Accepted Answer

Putting the dimensions for quantities in the expression containing

ε_{0}

.
From Coulomb's law, two stationary point charges

q_{1}

and

q_{2}

attract/repel each other with a force

F

which is directly proportional to the product of charges and inversely proportional to the square of distance

r

between them
That is,

F = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}}

\Rightarrow ε_{0} - \frac{1}{4 π} \frac{q _{1} q _{2}}{F r ^{2}}

∴

Dimensions of permittivity

ε_{0} = \frac{dimensions of q ^{2}}{dimensions of F \times dimensions of r ^{2}}

[ε_{0}] = \frac{[ A ^{2} T ^{2} ]}{[ M L T ^{- 2} ] [ L ^{2} ]} = [M^{- 1} L^{- 3} T^{4} A^{2}]

Q. Using mass $(M)$ , length $(L)$ , time $(T)$ and current $(A)$ as fundamental quantities, the dimension of permittivity is

$M L^{- 2} T^{2} A$

$M^{- 1} L^{- 3} T^{4} A^{2}$

$M L T^{- 2} A$

$M L^{2} T^{- 1} A^{2}$

Q. Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimension of permittivity is

ML−2T2A

M−1L−3T4A2

MLT−2A

ML2T−1A2

Q. Using mass $(M)$ , length $(L)$ , time $(T)$ and current $(A)$ as fundamental quantities, the dimension of permittivity is

$M L^{- 2} T^{2} A$

$M^{- 1} L^{- 3} T^{4} A^{2}$

$M L T^{- 2} A$

$M L^{2} T^{- 1} A^{2}$