A parallel plate capacitor of capacitance 50 is subjected to a...

The displacement current $I_d$ is equal to the conduction current $I_c$ during the charging of a capacitor, which is given by:
$I_d = C \dfrac{dV}{dt}$

During the interval $t = 0$ to $t = 2 \text{ ms}$, the voltage increases uniformly. The rate of change of voltage $\dfrac{dV}{dt}$ is the slope of the voltage-time graph:
$\dfrac{dV}{dt} = \dfrac{V_2 - V_1}{t_2 - t_1}$
$\dfrac{dV}{dt} = \dfrac{10 - 0}{2 \times 10^{-3} - 0} = \dfrac{10}{2 \times 10^{-3}} = 5000 \text{ V s}^{-1}$

Now, calculate the displacement current:
$I_d = (50 \times 10^{-6} \text{ F}) \times (5000 \text{ V s}^{-1})$
$I_d = 250000 \times 10^{-6} \text{ A} = 250 \times 10^{-3} \text{ A}$
$I_d = 250 \text{ mA}$

Note: Using the voltage value directly instead of its rate of change leads to the incorrect $500 \text{ } \mu\text{A}$, which represents the total charge stored, not the current.

Answer: $250 \text{ mA}$