Motion of Charge in Elliptical Orbit under Coulomb Field

To determine the correct statement(s), we analyze the motion of the charge $-q$ under the electrostatic attraction of the charge $+Q$:

1. Nature of the Force:
   The electrostatic force exerted by the charge $+Q$ (located at one focus of the elliptical orbit) on the charge $-q$ is given by Coulomb's law:
   $\vec{F} = -\frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2} \hat{r}$
   This is a central force, meaning it is always directed radially towards the position of $+Q$.

2. Angular Momentum:
   The torque $\vec{\tau}$ acting on the charge $-q$ about the position of charge $+Q$ is:
   $\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times \left(-\frac{Qq}{4\pi\varepsilon_0 r^3} \vec{r}\right) = \vec{0}$
   By the rotational analogue of Newton's second law:
   $\frac{d\vec{L}}{dt} = \vec{\tau} = \vec{0} \implies \vec{L} = \text{constant}$
   Therefore, the angular momentum of the charge $-q$ about $+Q$ remains constant in both magnitude and direction throughout its motion. Hence, statement (A) is correct.

3. Linear Momentum:
   Since a net attractive force $\vec{F} \neq \vec{0}$ acts on the charge $-q$, according to Newton's second law:
   $\frac{d\vec{p}}{dt} = \vec{F} \neq \vec{0}$
   The linear momentum $\vec{p}$ changes continuously in both magnitude and direction. Hence, statement (B) is incorrect.

4. Angular Velocity:
   The magnitude of angular momentum can be expressed as:
   $L = m r^2 \omega$
   In an elliptical orbit, the distance $r$ from the focus changes continuously. Since $L$ is constant:
   $\omega = \frac{L}{m r^2}$
   Because $r$ varies, the angular velocity $\omega$ must also vary. Hence, statement (C) is incorrect.

5. Linear Speed:
   The total mechanical energy of the system is conserved:
   $E = \frac{1}{2}mv^2 - \frac{Qq}{4\pi\varepsilon_0 r} = \text{constant}$
   As the distance $r$ varies, the potential energy changes, and consequently, the kinetic energy (and thus the linear speed $v$) must vary. The speed is greatest at the periapsis (closest approach) and least at the apoapsis (farthest approach). Hence, statement (D) is incorrect.

Conclusion:
The only correct statement is (A).