Dimension of epsilon-0 are -M-0 L-0 T-0 A-0 - -M-1- L-3- T-4 A-2 - -M-1- L-3- T-3 A-1 - -M-0 L-3- T-3 A-3 -

Question

Dimension of epsilon-0 are -M-0 L-0 T-0 A-0 -  -M-1- L-3- T-4 A-2 -  -M-1- L-3- T-3 A-1 -  -M-0 L-3- T-3 A-3 -

Toppr Admin · Accepted Answer

The electrostatic force is given as-F-Q1Q24-x3C0-x3B5-0R2 -x3B5-0-Q1Q24-x3C0-FR2 -Dimensions of -x3B5-0 are-A2T2MLT-x2212-2-xD7-L2 -A2T4ML3 -M-x2212-1L-x2212-3T4A2

	Column I (Physical quantity)		Column II (Dimensional formula)
(A)	Premittivity of free space	(p)	$[M^{0} L^{0} T^{- 1}]$
(B)	Radiant flux	(q)	$[M L^{3} T^{- 3} A^{- 2}]$
(C)	Resistivity	(r)	$[M L^{2} T^{- 3}]$
(D)	Hubble constant	(s)	$[M^{- 1} L^{- 3} T^{4} A^{2}]$

Dimension of $ϵ_{0}$ are
$[M^{0} L^{0} T^{0} A^{0}]$
$[M^{- 1} L^{- 3} T^{4} A^{2}]$
$[M^{- 1} L^{- 3} T^{3} A^{1}]$
$[M^{0} L^{- 3} T^{3} A^{3}]$

The electrostatic force is given as,
$F = \frac{Q_{1} Q_{2}}{4 π ε_{0} R^{2}}$

$ε_{0} = \frac{Q_{1} Q_{2}}{4 π F R^{2}}$

Dimensions of $ε_{0}$ are,

$= \frac{A^{2} T^{2}}{M L T^{- 2} \times L^{2}}$

$= \frac{A^{2} T^{4}}{M L^{3}}$

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

Dimension of ϵ0 are[M0L0T0A0][M−1L−3T4A2][M−1L−3T3A1][M0L−3T3A3]

The electrostatic force is given as,F=Q1Q24πε0R2 ε0=Q1Q24πFR2 Dimensions of ε0 are, =A2T2MLT−2×L2 =A2T4ML3 =M−1L−3T4A2

Dimension of $ϵ_{0}$ are
$[M^{0} L^{0} T^{0} A^{0}]$
$[M^{- 1} L^{- 3} T^{4} A^{2}]$
$[M^{- 1} L^{- 3} T^{3} A^{1}]$
$[M^{0} L^{- 3} T^{3} A^{3}]$

The electrostatic force is given as,
$F = \frac{Q_{1} Q_{2}}{4 π ε_{0} R^{2}}$

$ε_{0} = \frac{Q_{1} Q_{2}}{4 π F R^{2}}$

Dimensions of $ε_{0}$ are,

$= \frac{A^{2} T^{2}}{M L T^{- 2} \times L^{2}}$

$= \frac{A^{2} T^{4}}{M L^{3}}$

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