If $$\epsilon_0$$ denotes the permittivity of free space and $$\Phi_E$$ is the flux of the electric field through the area bounded by the closed surface, then dimension of $$\left(\epsilon_0 \frac{d\Phi_E}{dt}\right)$$ are that of :

To identify the physical quantity represented by $$\epsilon_0\dfrac{d\Phi_E}{dt}$$ we must compare its dimensions with those of the given options.

Step 1: Dimension of electric flux $$\Phi_E$$
Electric flux is defined as $$\Phi_E=\displaystyle\oint \mathbf E\cdot d\mathbf A$$.
• Dimension of electric field $$\mathbf E$$:
  $$\mathbf E=\dfrac{\text{Force}}{\text{Charge}}=\dfrac{MLT^{-2}}{IT}=MLT^{-3}I^{-1}$$.
• Area has dimension $$L^2$$.
Hence,
$$[\Phi_E]=MLT^{-3}I^{-1}\times L^2=ML^{3}T^{-3}I^{-1}$$.

Step 2: Dimension of $$\dfrac{d\Phi_E}{dt}$$
Time derivative multiplies the time dimension by $$T^{-1}$$:
$$\Bigl[\dfrac{d\Phi_E}{dt}\Bigr]=ML^{3}T^{-3}I^{-1}\times T^{-1}=ML^{3}T^{-4}I^{-1}.$$

Step 3: Dimension of $$\epsilon_0$$
From Coulomb’s law $$F=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2}$$ we get
$$\epsilon_0=\dfrac{q^2}{Fr^{2}}.$$
• $$[q]=IT,$$ so $$[q^2]=I^{2}T^{2}$$.
• $$[F]=MLT^{-2},\quad [r^2]=L^{2}.$$
Therefore,
$$[\epsilon_0]=\dfrac{I^{2}T^{2}}{MLT^{-2}\times L^{2}}=M^{-1}L^{-3}T^{4}I^{2}.$$

Step 4: Dimension of $$\epsilon_0\dfrac{d\Phi_E}{dt}$$
Multiply the results of Steps 2 and 3:
$$[\epsilon_0]\,[d\Phi_E/dt]=(M^{-1}L^{-3}T^{4}I^{2})(ML^{3}T^{-4}I^{-1})=I.$$

The final dimension is that of electric current. Thus $$\epsilon_0\dfrac{d\Phi_E}{dt}$$ represents a quantity having the dimensions of electric current.

Hence the correct choice is Option D: Electric current.