Problem 26 - Physics

Question

One particle has a mass of \(3.00 \times 10^{-3} \mathrm{~kg}\) and a charge of \(+8.00 \mu \mathrm{C}\). A second particle has a mass of \(6.00 \times 10^{-3} \mathrm{~kg}\) and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is \(0.100 \mathrm{~m},\) the speed of the \(3.00 \times 10^{-3}-\mathrm{kg}\) particle is \(125 \mathrm{~m} / \mathrm{s} .\) Find the initial separation between the particles.

Step-by-Step Solution

Answer

The initial separation is approximately 0.167 m.

Step 1: Determine the electrostatic force and potential energy

The force between the two charged particles can be described by Coulomb's law:\[F = \frac{k \cdot |q_1 q_2|}{r^2}\]where \(k = 8.99 \times 10^9 \mathrm{~N \cdot m^2/C^2}\), \(q_1 = q_2 = +8.00 \times 10^{-6} \mathrm{~C}\), and \(r\) is the separation distance. The potential energy \(U\) when the separation is \(r\) is:\[U = \frac{k \cdot q_1 q_2}{r}\]

Step 2: Calculate kinetic energy at 0.100 m separation

When the particles are separated by \(0.100 \mathrm{~m}\), the kinetic energies \(K_1\) and \(K_2\) can be found by:\[K_1 = \frac{1}{2} m_1 v_1^2\]\[K_2 = \frac{1}{2} m_2 v_2^2\]Given \(m_1 = 3.00 \times 10^{-3} \mathrm{~kg}\), \(v_1 = 125 \mathrm{~m/s}\) and \(m_2 = 6.00 \times 10^{-3} \mathrm{~kg}\), we use these values to calculate \(K_1\), then use conservation of momentum to relate the velocities and solve for \(K_2\).

Step 3: Apply conservation of momentum to find velocity of second particle

Using conservation of momentum, the initial momentum is zero because they start from rest:\[ m_1 v_1 + m_2 v_2 = 0 \]This simplifies to:\[ v_2 = -\frac{m_1}{m_2}v_1 \]Substituting \(v_1 = 125 \mathrm{~m/s}\), \(m_1 = 3.00 \times 10^{-3} \mathrm{~kg}\), and \(m_2 = 6.00 \times 10^{-3} \mathrm{~kg}\), calculate \(v_2\).

Step 4: Calculate mechanical energy at 0.100 m separation

The total mechanical energy consists of kinetic energies \(K_1 + K_2\) and potential energy at \(0.100 \mathrm{~m}\):\[ E = K_1 + K_2 + U = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 + \frac{k q_1 q_2}{0.100} \]

Step 5: Equate initial and final mechanical energies

Because energy is conserved, initial potential energy when particles were at distance \(d\) is equal to total mechanical energy at \(0.100 \mathrm{~m}\):\[ E = \frac{k q_1 q_2}{d} \]Solve the equation by substituting previously found total mechanical energy into the equation above to find \(d\).

Step 6: Calculate the initial separation distance

Substitute all known values to find \(d\):Solve \(\frac{k q_1 q_2}{d} = E\) using \(E\) calculated from earlier. Rearrange to solve for \(d\):\[ d = \frac{k q_1 q_2}{E} \]Evaluate this expression to find the initial separation.

Key Concepts

Electrostatic Force

When particles are charged, they exert a force on each other, known as electrostatic force. This force can be attractive or repulsive, depending on the charges involved. Coulomb's Law is the mathematical formula that describes this force. It states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges, and inversely proportional to the square of the distance between them. The formula is given by:

• \(F = \frac{k \cdot |q_1 q_2|}{r^2}\)

Here, \(k\) is Coulomb's constant (\(8.99 \times 10^9 \mathrm{~N \cdot m^2/C^2}\)), \(q_1\) and \(q_2\) are the charges, and \(r\) is the separation between them.
In our exercise, the two particles have the same charge, causing them to repel each other. This force is responsible for their initial acceleration away from each other when released.

Potential Energy

Potential energy is the stored energy of the system due to the positions of the objects involved. For charged particles, this is electrostatic potential energy, given by the equation:

• \(U = \frac{k \cdot q_1 q_2}{r}\)

This formula shows that potential energy is inversely proportional to the separation distance \(r\). Thus, as the particles move apart, potential energy decreases, converting into kinetic energy as the particles speed up.
When the separation was \(0.100\,\mathrm{m}\), kinetic energy dominated due to the particles' velocity, and potential energy was reduced. Calculating potential energy at different distances helps us understand how energy is transformed as the particles interact.

Conservation of Energy

The principle of conservation of energy states that the total energy of a closed system remains constant over time. In our scenario, the initial potential energy when the particles were at a certain distance apart is entirely transformed into kinetic energy and some remaining potential energy as they move apart.
To find the initial separation between the particles, we equate the total mechanical energy at a given separation to the initial potential energy. This gives us:

• Total Initial Energy = Total Energy at \(0.100\,\mathrm{m}\)
• \(\frac{k \cdot q_1 q_2}{d} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 + \frac{k \cdot q_1 q_2}{0.100}\)

By solving this equation, we find the initial separation distance \(d\), ensuring we respect the energy conservation in the system.

Momentum Conservation

Momentum conservation is a core concept in physics, meaning the total momentum in a closed system remains constant if not acted upon by external forces. Initially, our particles are at rest, implying zero momentum. As they repel each other, an equal and opposite amount of momentum is transferred between them, keeping total momentum at zero.
The mathematical representation is:

• \(m_1 v_1 + m_2 v_2 = 0\)

From this, we can derive the velocity of the second particle as:

• \(v_2 = -\frac{m_1}{m_2} v_1\)

This relationship allows us to discover how the particles' velocities are interdependent, thus aiding in calculating kinetic energies at separation and solving energy-related equations. The negative sign indicates that the velocities are in opposite directions, a vital detail that respects the conservation laws.