Understanding Charge Motion in an Electric Field

This problem involves a charged particle orbiting another fixed charged particle. The force responsible for keeping the orbiting particle in a circular path is the electrostatic force between the two charges. This electrostatic force acts as the centripetal force required for circular motion.

Let's identify the key elements:

• Fixed charge: $+q_2$
• Orbiting charge: $-q_1$
• Mass of orbiting particle: $m$
• Radius of circular path: $r$
• Period of revolution: $T$

Analyzing the Forces at Play

The force between the two charges $+q_2$ and $-q_1$ is attractive. The magnitude of this electrostatic force is given by Coulomb's Law:

\(F_e = \frac{1}{4\pi \epsilon_0} \frac{|(+q_2)(-q_1)|}{r^2}\)

Since $|(+q_2)(-q_1)| = q_1 q_2$, the magnitude of the electrostatic force is:

\(F_e = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}\)

For the charge $-q_1$ to move in a circular path of radius $r$ with speed $v$, a centripetal force is required. The magnitude of the centripetal force is given by:

\(F_c = \frac{mv^2}{r}\)

Setting up the Equation of Motion

In this scenario, the electrostatic force provides the necessary centripetal force. Therefore, we can equate the two forces:

\(F_e = F_c\)

\(\frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2} = \frac{mv^2}{r}\)

Now, we need to solve for the speed \(v\) of the orbiting charge:

\(v^2 = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2} \times \frac{r}{m}\)

\(v^2 = \frac{q_1 q_2}{4\pi \epsilon_0 m r}\)

\(v = \sqrt{\frac{q_1 q_2}{4\pi \epsilon_0 m r}}\)

Calculating the Period of Revolution

The period of revolution \(T\) is the time it takes for the charge to complete one full circle. The distance covered in one revolution is the circumference of the circle, \(2\pi r\). The speed \(v\) is related to the period \(T\) and radius \(r\) by:

\(v = \frac{\text{Distance}}{\text{Time}} = \frac{2\pi r}{T}\)

We can rearrange this equation to solve for \(T\):

\(T = \frac{2\pi r}{v}\)

Now, substitute the expression for \(v\) that we found earlier:

\(T = \frac{2\pi r}{\sqrt{\frac{q_1 q_2}{4\pi \epsilon_0 m r}}}\)

To simplify, we can move \(2\pi r\) inside the square root by squaring it:

\(T = \sqrt{(2\pi r)^2 \times \frac{4\pi \epsilon_0 m r}{q_1 q_2}}\)

\(T = \sqrt{4\pi^2 r^2 \times \frac{4\pi \epsilon_0 m r}{q_1 q_2}}\)

\(T = \sqrt{\frac{4\pi^2 r^2 \cdot 4\pi \epsilon_0 m r}{q_1 q_2}}\)

\(T = \sqrt{\frac{16\pi^3 r^3 \epsilon_0 m}{q_1 q_2}}\)

\(T = \sqrt{\frac{16\pi^3 \epsilon_0 m r^3}{q_1 q_2}}\)

Comparing with Given Options

Let's compare our derived expression for the period \(T\) with the given options:

• Option 1: \(\sqrt {\frac{{16{\pi ^3}{ \in _0}m{r^3}}}{{{q_1}}}} \)
• Option 2: \(\sqrt {\frac{{16{\pi ^2}{ \in _0}m{r^3}}}{{{q_1}}}} \)
• Option 3: \(\sqrt {\frac{{16{\pi ^3}{ \in _0}m{r^3}}}{{{q_1\space q_2}}}} \)
• Option 4: \(\sqrt {\frac{{16{\pi ^2}{ \in _0}m{r^3}}}{{{q_1\space q_2}}}} \)

Our derived period \(T = \sqrt{\frac{16\pi^3 \epsilon_0 m r^3}{q_1 q_2}}\) exactly matches Option 3.

Revision Table: Key Concepts and Formulas

Additional Information on Circular Motion and Electrostatics

This problem is analogous to planetary motion, where the gravitational force provides the centripetal force. Here, the electrostatic force plays that role. The orbiting charge has kinetic energy due to its motion and potential energy due to its position in the electric field of the fixed charge. The total energy (kinetic + potential) is conserved in the absence of non-conservative forces.

• Electric Potential Energy: The potential energy \(U\) of charge \(-q_1\) in the electric field of charge \(+q_2\) at distance \(r\) is given by \(U = \frac{1}{4\pi \epsilon_0} \frac{(+q_2)(-q_1)}{r} = - \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r}\). Note the negative sign, indicating an attractive potential.
• Kinetic Energy: The kinetic energy \(K\) of the orbiting charge is \(K = \frac{1}{2}mv^2\). Using the expression for \(v^2\) we derived: \(K = \frac{1}{2}m \left(\frac{q_1 q_2}{4\pi \epsilon_0 m r}\right) = \frac{q_1 q_2}{8\pi \epsilon_0 r}\).
• Total Energy: The total energy \(E = K + U = \frac{q_1 q_2}{8\pi \epsilon_0 r} - \frac{q_1 q_2}{4\pi \epsilon_0 r} = - \frac{q_1 q_2}{8\pi \epsilon_0 r}\). The negative total energy indicates that the charge is bound to the fixed charge.
• Angular Velocity: The angular velocity \(\omega\) is related to the speed \(v\) and radius \(r\) by \(v = r\omega\). The period \(T\) is related to angular velocity by \(T = \frac{2\pi}{\omega}\).

Understanding the interplay between forces, motion, and energy is crucial for solving such problems in electrostatics and mechanics.