A steady conduction current I is used to charge a parallel plate...

The electric field $E$ between the plates of a parallel plate capacitor is given by $E = \dfrac{q}{\epsilon_0 A}$, where $q$ is the charge on the capacitor plates and $A$ is the area of the plates.

Differentiating both sides with respect to time $t$, we get:

$\dfrac{dE}{dt} = \dfrac{1}{\epsilon_0 A} \dfrac{dq}{dt}$

The rate of change of charge on the plates is equal to the steady conduction current $I$, so $\dfrac{dq}{dt} = I$.

Substituting this into the equation, we obtain:

$\dfrac{dE}{dt} = \dfrac{I}{\epsilon_0 A}$

Alternatively, using the concept of displacement current $I_d$, we know that the displacement current in the gap is equal to the conduction current $I$ in the connecting wires.

$I_d = \epsilon_0 A \dfrac{dE}{dt}$

Since $I_d = I$, we have:

$I = \epsilon_0 A \dfrac{dE}{dt} \Rightarrow \dfrac{dE}{dt} = \dfrac{I}{\epsilon_0 A}$

Answer: $\dfrac{I}{\epsilon_0 A}$