91Ó°ÊÓ

A uniform electric field is a region where the electric force experienced by a charge is constant both in magnitude and direction. This means that for any point in this field, a charge will feel the same push or pull, making calculations more straightforward as the conditions are consistent throughout. In our exercise, the particles are in a uniform electric field of \(+2500 \, \mathrm{N/C}\). This value tells us how much force a \(1 \, \mu \mathrm{C}\) charge will experience at any given point in this field.
Understanding how charges interact in such a field lays a foundation for grasping more complex scenarios where fields might not be uniform. When particles with different charges enter this uniform electric field, each experiences a force that is proportional to its charge. This principle is crucial in this exercise, as each particle's force is calculated based on its own charge and the consistent field strength.

Electric forces are the push or pull that charged objects exert on each other due to their electric fields. This force can either be attractive or repulsive, depending on the signs of the interacting charges.
In our exercise, each particle experiences an electric force due to the uniform electric field. The formula \( F = qE \) gives us the electric force, where \( q \) is the charge of the particle and \( E \) is the electric field strength. Therefore:

• For particle 1 with a charge of \(-7.0 \, \mu \mathrm{C}\), the force is calculated as \( F_1 = -7.0 \times 10^{-6} \times 2500 \).
• For particle 2 with a charge of \(+18 \, \mu \mathrm{C}\), the force is \( F_2 = +18 \times 10^{-6} \times 2500 \).

These forces, applied in opposite directions, cause the charges to accelerate along the field.

Newton's Second Law tells us that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass, expressed as \( F = ma \). This law is applied to find the accelerations of the particles due to the electric forces witnessed in the uniform electric field.
For the two particles:

• Particle 1 has an acceleration \( a_1 = \frac{F_1}{m_1} \).
• Particle 2 has an acceleration \( a_2 = \frac{F_2}{m_2} \).

Even though these particles have different masses and charges, they need to have the same acceleration to maintain a constant distance, \( d \), between them. Consequently, we set \( a_1 = a_2 \) to find this relationship.

Coulomb's Law provides the formula to calculate the force of electrostatic interaction between two point charges. This law is especially useful when considering forces between charged particles that bring them to equilibrium.
It is expressed as: \[ F = \frac{k |q_1 q_2|}{d^2} \]where \( k \) is the electrostatic constant \( 8.99 \times 10^9 \, \mathrm{Nm^2/C^2} \), \( q_1 \) and \( q_2 \) are the charges, and \( d \) is the separation distance.
In this exercise, the electrostatic force between the particles balances the electric forces due to the field, keeping \( d \) constant. Solving for \( d \) requires setting the inward electric force \( F_1 \) equal to the electrostatic repelling force due to the other particle's field, leading to \( F_1 + F_{12(\text{electrostatic})} = 0 \). This balance permits us to find \( d \), keeping in mind all forces must cancel each other out for a steady state.