Let denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then: [2012]

Question

Let $[ε_{0}]$ denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then: [2012]

Smart Achievers · Accepted Answer

(2)

$As we know, F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{R^{2}} \Rightarrow ε_{0} = \frac{q_{1} q_{2}}{4 π F R^{2}}$

$Hence, ε_{0} = \frac{C^{2}}{N \cdot m^{2}} = \frac{{[A T]}^{2}}{[M L T^{- 2}] [L^{2}]} = [M^{- 1} L^{- 3} T^{4} A^{2}]$

Solution

Q.
Let $[ε_{0}]$ denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then: [2012]

1 $ε_{0} = [M^{- 1} L^{- 3} T^{2} A]$

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

3 $ε_{0} = [M^{1} L^{2} T^{1} A^{2}]$

4 $ε_{0} = [M^{1} L^{2} T^{1} A]$

Ans.
(2)

$As we know, F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{R^{2}} \Rightarrow ε_{0} = \frac{q_{1} q_{2}}{4 π F R^{2}}$

$Hence, ε_{0} = \frac{C^{2}}{N \cdot m^{2}} = \frac{{[A T]}^{2}}{[M L T^{- 2}] [L^{2}]} = [M^{- 1} L^{- 3} T^{4} A^{2}]$

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Solution

Q. Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then: [2012]

1 ε0=[M-1L-3T2A]

2 ε0=[M-1L-3T4A2]

3 ε0=[M1L2T1A2]

4 ε0=[M1L2T1A]

Ans. (2) As we know, F=14πε0q1q2R2 ⇒ ε0=q1q24πFR2 Hence, ε0=C2N·m2=[AT]2[MLT-2][L2]=[M-1L-3T4A2]