A physical quantity $\vec{S}$ is defined as $\vec{S} = \frac{\vec{E} \times \vec{B}}{\mu_0}$, where $\vec{E}$ is electric field, $\vec{B}$ is magnetic field and $\mu_0$ is the permeability of free space. The dimensions of $\vec{S}$ are the same as the dimensions of which of the following quantities?

Step 1: Find the dimensions of electric field, magnetic field, and permeability.

The dimensions of electric field $E$ are derived from force per unit charge.

$$[E] = \frac{[F]}{[q]}$$ $$[E] = \text{MLT}^{-3}\text{A}^{-1}$$

The dimensions of magnetic field $B$ are derived from the magnetic force equation.

$$[B] = \frac{[F]}{[q][v]}$$ $$[B] = \text{MT}^{-2}\text{A}^{-1}$$

The dimensions of permeability of free space $\mu_0$ are derived from the force per unit length between two parallel currents.

$$[\mu_0] = \frac{[F]}{[I]^2}$$ $$[\mu_0] = \text{MLT}^{-2}\text{A}^{-2}$$

Step 2: Calculate the dimensions of $\vec{S}$.

Substitute the dimensions into the expression for $\vec{S}$.

$$[S] = \frac{[E][B]}{[\mu_0]}$$ $$[S] = \frac{(\text{MLT}^{-3}\text{A}^{-1})(\text{MT}^{-2}\text{A}^{-1})}{\text{MLT}^{-2}\text{A}^{-2}}$$ $$[S] = \text{MT}^{-3}$$

Step 3: Evaluate the dimensions of the options.

Evaluate Option (1).

$$\left[\frac{\text{Energy}}{\text{Charge} \times \text{Current}}\right] = \frac{\text{ML}^2\text{T}^{-2}}{(\text{AT})(\text{A})}$$ $$\left[\frac{\text{Energy}}{\text{Charge} \times \text{Current}}\right] = \text{ML}^2\text{T}^{-3}\text{A}^{-2}$$

Evaluate Option (2).

$$\left[\frac{\text{Force}}{\text{Length} \times \text{Time}}\right] = \frac{\text{MLT}^{-2}}{\text{LT}}$$ $$\left[\frac{\text{Force}}{\text{Length} \times \text{Time}}\right] = \text{MT}^{-3}$$

Evaluate Option (3).

$$\left[\frac{\text{Energy}}{\text{Volume}}\right] = \frac{\text{ML}^2\text{T}^{-2}}{\text{L}^3}$$ $$\left[\frac{\text{Energy}}{\text{Volume}}\right] = \text{ML}^{-1}\text{T}^{-2}$$

Evaluate Option (4).

$$\left[\frac{\text{Power}}{\text{Area}}\right] = \frac{\text{ML}^2\text{T}^{-3}}{\text{L}^2}$$ $$\left[\frac{\text{Power}}{\text{Area}}\right] = \text{MT}^{-3}$$

Step 4: Identify the matching physical quantity.

The quantity $\vec{S}$ represents the Poynting vector, which physically describes the energy flux density (intensity).

$$S = \frac{\text{Power}}{\text{Area}}$$

Answer: $\dfrac{\text{Power}}{\text{Area}}$