`(E^(2))/(mu_(0))` has the dimensions (`E=` electric field, `mu_(0)=` permeabililty of free space)

Options

A. `[M^(2)L^(3)T^(-2)A^(2)]`
B. `[MLT^(-4)]`
C. `[ML^(3)T^(-2)]`
D. `[M^(-1)L^(2)TA^(-2)]`

Answer

To find the dimensions of the expression \(\frac{E^2}{\mu_0}\), where \(E\) is the electric field and \(\mu_0\) is the permeability of free space, we can follow these steps:

### Step 1: Identify the dimensions of the electric field \(E\)
The electric field \(E\) is defined as force per unit charge. The dimensions of force \(F\) are given by:
\[ [F] = [M][L][T^{-2}] \]
where \(M\) is mass, \(L\) is length, and \(T\) is time. The dimensions of charge \(Q\) can be represented as \(Q\).

Thus, the dimensions of the electric field \(E\) can be expressed as:
\[ [E] = \frac{[F]}{[Q]} = \frac{[M][L][T^{-2}]}{[Q]} = [M][L][T^{-2}][Q^{-1}] \]

### Step 2: Calculate the dimensions of \(E^2\)
Now, squaring the dimensions of \(E\):
\[ [E^2] = \left([M][L][T^{-2}][Q^{-1}]\right)^2 = [M^2][L^2][T^{-4}][Q^{-2}] \]

### Step 3: Identify the dimensions of \(\mu_0\)
The permeability of free space \(\mu_0\) can be related to the speed of light \(c\) and the permittivity of free space \(\epsilon_0\) by the equation:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]
From this, we can express \(\mu_0\) as:
\[ \mu_0 = \frac{1}{c^2 \epsilon_0} \]
The dimensions of \(\epsilon_0\) can be derived from Coulomb's law:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
Rearranging gives:
\[ \epsilon_0 = \frac{q_1 q_2}{4\pi F r^2} \]
Thus, the dimensions of \(\epsilon_0\) are:
\[ [\epsilon_0] = \frac{[Q^2]}{[M][L^2][T^{-2}]} = [M^{-1}][L^{-3}][T^4][Q^2] \]
Now substituting this into the expression for \(\mu_0\):
\[ [\mu_0] = \frac{1}{c^2 [\epsilon_0]} = \frac{1}{[L^2][T^{-2}]} \cdot [M][L^3][T^{-4}][Q^{-2}] = [M][L^{-1}][T^2][Q^2] \]

### Step 4: Calculate the dimensions of \(\frac{E^2}{\mu_0}\)
Now we can find the dimensions of \(\frac{E^2}{\mu_0}\):
\[ \left[\frac{E^2}{\mu_0}\right] = \frac{[E^2]}{[\mu_0]} = \frac{[M^2][L^2][T^{-4}][Q^{-2}]}{[M][L^{-1}][T^2][Q^2]} \]
This simplifies to:
\[ = [M^{2-1}][L^{2+1}][T^{-4-2}][Q^{-2-2}] = [M^1][L^3][T^{-6}][Q^{-4}] \]

### Final Result
Thus, the dimensions of \(\frac{E^2}{\mu_0}\) are:
\[ [M][L^3][T^{-6}][Q^{-4}] \]