Let [epsi(0)] denote the dimensional for...

Let `[epsi_(0)]` denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T= time and A= electric current, then

`[epsi_(0)]=[M^(-1)L^(2)T^(-1)A]`

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

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

`[epsi_(0)]=[M^(-1)L^(2)T^(-1)A^(-2)]`

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To find the dimensional formula of the permittivity of vacuum (denoted as \(\epsilon_0\)), we can use Coulomb's law, which relates the force between two point charges to the distance between them. Here’s a step-by-step solution: ### Step 1: Write down Coulomb's Law According to Coulomb's law, the force \(F\) between two charges \(Q_1\) and \(Q_2\) separated by a distance \(r\) is given by: \[ F = \frac{1}{4\pi \epsilon_0} \frac{Q_1 Q_2}{r^2} \] ### Step 2: Rearrange the equation to find \(\epsilon_0\) From the equation, we can rearrange it to express \(\epsilon_0\): \[ \epsilon_0 = \frac{1}{4\pi} \frac{Q_1 Q_2}{F r^2} \] ### Step 3: Substitute the dimensions of the quantities - The dimension of force \(F\) is given by: \[ [F] = M L T^{-2} \] - The charge \(Q\) can be expressed in terms of current \(I\) and time \(t\): \[ Q = I \cdot t \] Thus, the dimension of charge \(Q\) is: \[ [Q] = A T \] ### Step 4: Substitute \(Q_1\) and \(Q_2\) into the equation Substituting \(Q_1\) and \(Q_2\) into the equation for \(\epsilon_0\): \[ Q_1 Q_2 = (A T)(A T) = A^2 T^2 \] ### Step 5: Substitute everything back into the equation for \(\epsilon_0\) Now substituting the dimensions back into the equation for \(\epsilon_0\): \[ \epsilon_0 = \frac{1}{4\pi} \frac{A^2 T^2}{M L T^{-2} L^2} \] ### Step 6: Simplify the expression This simplifies to: \[ \epsilon_0 = \frac{A^2 T^2}{M L T^{-2} L^2} = \frac{A^2 T^2}{M L^3 T^{-2}} = \frac{A^2 T^4}{M L^3} \] ### Step 7: Write the final dimensional formula Thus, the dimensional formula for the permittivity of vacuum \(\epsilon_0\) is: \[ [\epsilon_0] = M^{-1} L^{-3} T^{4} A^{2} \] ### Conclusion The dimensional formula of the permittivity of vacuum is: \[ [\epsilon_0] = M^{-1} L^{-3} T^{4} A^{2} \]

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Let [in_(0)] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = elctric current , then :

(a) `in_(0) = [M^(-1)L^(-3)T^(2)A]`

(b) `in_(0) = [M^(1)L^(3)T^(5)A^(2)]`

( c ) `in_(0) = [M^(1)L^(2)T^(1)A^(2)]`

(d) `in_(0) = [M^(1)L^(-2)T^(1)A]`

Let [in_0] denote the permittivity of vaccum. If M = mass, L = length, T = time and A = electric current, then :

`[in_0]= M^(-1) L^(-3) T^2 A`

`[in_0] = M^(-1) L^(-3) T^4 A^2`

`[in_0] = MLT^(-2) A(-2)`

`[in_0] = ML^2 T^(-1)A`

Let [epsi_(0)] denote the dimensional formula of the permittivity of the vacuum and [mu_(0)] that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current :

`[epsi_(0)]=[M^(-1)L^(-3)T^(2)I]`

`[epsi_(0)]=[M^(-1)L^(-3)T^(4)I^(2)]`

`[mu_(0)]=[MLT^(-2)I^(-2)]`

`[mu_(0)]=[ML^(2)T^(-1)I]`