Animated Solution for Physics - Electrostatics: Under the influence of the coulomb field of a fixed charge $+Q$, a charge $-q$ is moving around it in an elliptical orbit. Find out the correct statement(s).

Imagine a universe where a single negative charge, $-q$, dances around a massive, fixed positive charge, $+Q$. This isn't just a random dance; it's a perfectly choreographed elliptical orbit dictated by the laws of electrostatics. The only force acting on our moving charge is the Coulomb attraction.

This force is incredibly special—it is a central force. A central force is one that always points directly towards or away from a fixed point (in this case, the charge $+Q$). Mathematically, we can write this force as:

Now, let's look at the rotational dynamics of this system. Torque ($\vec{\tau }$) is the rotational equivalent of force, and it tells us how much a force causes an object to rotate about a specific point. It is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:

Here is where the magic happens. Because the electrostatic force is a central force, it always lies exactly along the line of the position vector $\vec{r}$. The angle $\theta$ between $\vec{r}$ and ${\vec{F}}_e$ is exactly $180^{\circ }$.

Since the cross product depends on the sine of the angle between the vectors, and $\sin (180^{\circ })=0$, the net torque acting on the charge $-q$ about the fixed charge $+Q$ is absolutely zero!

Newton's second law for rotation states that the net torque on a system is equal to the rate of change of its angular momentum ($\vec{L}$):

Since we just proved that $\vec{\tau }=0$, it immediately follows that:

This is a profound result. In any central force field, angular momentum is strictly conserved.

What about the other options? In an elliptical orbit, the distance $r$ between the charges is constantly changing. As $-q$ gets closer to $+Q$, it speeds up (converting potential energy into kinetic energy), and as it moves further away, it slows down.

Because the speed $v$ and the direction of motion change, the linear momentum ($\vec{p}=m\vec{v}$) and the linear speed ($v$) are not constant. Furthermore, since angular momentum $L=m{r}^2\omega$ is constant but $r$ changes, the angular velocity ($\omega$) must also change to compensate. Thus, only the angular momentum remains a steadfast constant in this beautiful orbital dance.