Problem 12 - Fundamentals of Physics

Question

Two particles are fixed on an \(x\) axis. Particle 1 of charge \(40 \mu \mathrm{C}\) is located at \(x=-2.0 \mathrm{~cm} ;\) particle 2 of charge \(Q\) is located at \(x=3.0 \mathrm{~cm} .\) Particle 3 of charge magnitude \(20 \mu \mathrm{C}\) is released from rest on the \(y\) axis at \(y=2.0 \mathrm{~cm} .\) What is the value of \(Q\) if the initial acceleration of particle 3 is in the positive direction of (a) the \(x\) axis and (b) the \(y\) axis?

Step-by-Step Solution

Answer

Solve for \( Q \) by ensuring net forces are positive in each axis direction through force component analysis.

Step 1: Understand the Problem

We need to find the charge \( Q \) such that the initial acceleration of particle 3 is positive along the \( x \) axis and separately along the \( y \) axis. Particle 3 is placed on the \( y \) axis and experiences forces due to particles 1 and 2 on the \( x \) axis.

Step 2: Calculate Distance from Charges to Particle 3

Distances from particle 3 to particles 1 and 2 are required: For particle 1, located at \((-2.0, 0)\), to \((0, 2.0)\): \[ r_1 = \sqrt{(-2.0)^2 + (2.0)^2} = \sqrt{8} \text{ cm} \]For particle 2, located at \((3.0, 0)\), to \((0, 2.0)\): \[ r_2 = \sqrt{(3.0)^2 + (2.0)^2} = \sqrt{13} \text{ cm} \]

Step 3: Calculate Force Components on Particle 3

Using Coulomb's law, the magnitude of force exerted by a charge is given by \[ F_{12} = \frac{k |q_1| |q_3|}{r^2} \] where \( k \approx 8.99 \times 10^9 \) N\(\cdot\)m²/C². Particle 3 feels forces from particles 1 and 2, which can be decomposed into their \( x \) and \( y \) components.

Step 4: Initial Acceleration in Positive x-axis Direction

For positive \( x \) acceleration:The \( x \)-component of the total force should be positive. Particle 3's \( x \)-component acceleration relative to particle 1 (as a reference) is \[ F_{1x} = \frac{k \, q_1 \, q_3 \, (-2/\sqrt{8})}{8} \] Similarly for \( P_2 \):\[ F_{2x} = \frac{k \, q_2 \, q_3 \, (3/\sqrt{13})}{13} \]Set the net \( F_{net_x} > 0 \) and solve \[ \frac{k \, q_1 \, q_3 (-2/\sqrt{8})}{8} + \frac{k \, Q \, q_3 (3/\sqrt{13})}{13} > 0 \] to find the limit for \( Q \).

Step 5: Initial Acceleration in Positive y-axis Direction

For positive \( y \) acceleration:The net force should be positive in the \( y \):*From particle 1*: \[ F_{1y} = \frac{k \, q_1 \, q_3 \, (2/\sqrt{8})}{8} \] *From particle 2*: \[ F_{2y} = \frac{k \, Q \, q_3 (-2/\sqrt{13})}{13} \]Set net \( F_{net_y} > 0 \) and solve\[ \frac{k \, q_1 \, q_3 \, (2/\sqrt{8})}{8} + \frac{k \, Q \, q_3 (-2/\sqrt{13})}{13} > 0 \] to find the required \( Q \).

Step 6: Conclusion

The solutions for each part \( Q \) should fulfill conditions specific to each directional force and then analyze which direction \( Q \) must have to align with the force expectations.

Key Concepts

Electric Charge

An electric charge is a fundamental property of matter responsible for electric forces and interactions. Charges can be positive or negative.They exert forces on each other, either attracting if opposite or repelling if identical.
In the given example, we have three particles with different charges and positions:

• Particle 1 has a charge of \(40 \mu \mathrm{C}\) and is located at \(x=-2.0 \mathrm{~cm}\).
• Particle 2 has an unknown charge \(Q\) located at \(x=3.0 \mathrm{~cm}\).
• Particle 3, with a charge of \(20 \mu \mathrm{C}\), is on the \(y\) axis at \(y=2.0 \mathrm{~cm}\).

The charges on these particles create electric fields, leading to forces acting on each other, crucial for the dynamics described.

Force Components

In physics, forces have components, especially in multidimensional spaces like 2D or 3D. Here, we're in a 2D plane, dealing with the forces on particle 3 caused by particles 1 and 2.Using Coulomb's Law, the forces calculated are vector quantities.
Coulomb's Law states:\[F = \frac{k \cdot |q_1| \cdot |q_2|}{r^2}\]where

• \(F\) is the force between two charges,
• \(k\) is Coulomb's constant (\(8.99 \times 10^9 \) N·m²/C²),
• \(q_1\) and \(q_2\) are charges, and \(r\) is the distance between charges.

The forces experienced by particle 3 are split into components along the \(x\) and \(y\) axes. For each influence of particles 1 and 2, decompose as follows:

• \(F_{x_1}, F_{x_2}\) for \(x\) components from particles 1 and 2.
• \(F_{y_1}, F_{y_2}\) for \(y\) components from particles 1 and 2.

These components help to independently analyze particle 3's acceleration along the \(x\) and \(y\) directions.

Initial Acceleration

Acceleration is the rate of change of velocity, influenced by the net force acting on a particle. The problem highlights the initial acceleration of particle 3 due to forces exerted by particles 1 and 2.
Newton's Second Law, \(F = ma\), connects net force \(F\) with mass \(m\) and acceleration \(a\).For particle 3:

• The goal is to have its acceleration component be positive along the \(x\) axis or \(y\) axis, separately.
• The expression leveraging force components follows \(F_{net} = ma_{net}\).

By breaking forces into \(x\) and \(y\) components separately and equating them for a positive direction, the required charge \(Q\) providing this acceleration is deduced.

Two-Body Problem

The scenario is a simplified version of the two-body problem, where the effect of multiple bodies' gravitation, or here, electric interaction is considered. Instead of simplifying it as pair interactions, each body influences the other's motion.Particle 3 experiences forces from both particles 1 and 2. To solve for \(Q\), one needs to simultaneously consider the influences from both fixed particles, especially:

• The net forces acting horizontally and vertically due to each charge.
• Their relative positions, calculated using Pythagorean theorem to understand distances.

This technique simplifies initial complex interactions into manageable components, helping us determine the necessary condition on \(Q\) to drive motion in the required directions.