We can use Gauss’s law to analyze the dimensions. Gauss’s law states that

$$ \Phi_E = \frac{q}{\epsilon_0}, $$

where

$$\Phi_E$$ is the electric flux,

$$q$$ is the electric charge,

$$\epsilon_0$$ is the permittivity of free space.

To find the dimensions of $$\epsilon_0 \frac{d\Phi_E}{dt}$$, follow these steps:

Differentiate Gauss’s law with respect to time:

$$ \frac{d\Phi_E}{dt} = \frac{1}{\epsilon_0}\frac{dq}{dt}. $$

Multiply the equation by $$\epsilon_0$$:

$$ \epsilon_0 \frac{d\Phi_E}{dt} = \frac{dq}{dt}. $$

Recognize that the time rate of change of charge, $$\frac{dq}{dt}$$, is by definition the electric current.

Thus, the dimensions of $$\epsilon_0 \frac{d\Phi_E}{dt}$$ are those of electric current.

The correct answer is: Option C, electric current.