A point charge travels along a straight line with a constant...

The electric field at a distance $x$ from the point charge is given by Coulomb's law:

$E = \dfrac{q}{4\pi\epsilon_0 x^2}$

Since the area $A$ is small and oriented perpendicular to the direction of motion, the electric field is uniform over the area and parallel to the area vector. The electric flux through the area $A$ is:

$\Phi_E = E A = \dfrac{q A}{4\pi\epsilon_0 x^2}$

The displacement current $I_d$ is defined as:

$I_d = \epsilon_0 \dfrac{d\Phi_E}{dt}$

Differentiating the electric flux with respect to time, we get:

$\dfrac{d\Phi_E}{dt} = \dfrac{d}{dt} \left( \dfrac{q A}{4\pi\epsilon_0 x^2} \right) = \dfrac{q A}{4\pi\epsilon_0} \left( -\dfrac{2}{x^3} \right) \dfrac{dx}{dt}$

Since the charge is moving towards the area with a constant velocity $v$, the distance $x$ is decreasing at the rate of $v$. Therefore, the rate of change of distance is:

$\dfrac{dx}{dt} = -v$

Substituting this into the derivative of flux:

$\dfrac{d\Phi_E}{dt} = \dfrac{q A}{4\pi\epsilon_0} \left( -\dfrac{2}{x^3} \right) (-v) = \dfrac{q A v}{2\pi\epsilon_0 x^3}$

Finally, the displacement current is:

$I_d = \epsilon_0 \left( \dfrac{q A v}{2\pi\epsilon_0 x^3} \right) = \dfrac{q A v}{2\pi x^3}$