Electric Field Between Parallel Plates Calculating Particle Motion

Hey guys! Let's dive into an exciting physics problem involving electric fields and charged particles. We're going to break down a scenario with two parallel metal plates, explore the electric field they create, and see how a charged particle behaves within that field. So, buckle up and let's get started!

The Setup: Charged Parallel Plates

Imagine we have two parallel metal plates, like giant sheets of foil, positioned a short distance apart. These plates are electrically charged, but with opposite signs – one positive and one negative. This difference in charge creates an electric field between the plates. Think of it like an invisible force field that affects charged particles.

In our specific problem, these plates are separated by 10 cm (which is 0.1 meters, remember to use the right units!). The electric field generated between them is quite strong, measuring 6 x 10^7 N/C (Newtons per Coulomb). This tells us the force that would be exerted on a unit positive charge placed in this field. Now, we introduce a small charged particle – a point charge – with a charge of 2 x 10^-6 C (Coulombs) and a mass of 5 x 10^-6 kg (kilograms). This little guy is placed right on the positive plate. What happens next? That's what we're going to explore!

To really understand what's going on, let's break down the key concepts. First, we have the electric field. An electric field is a region of space where an electric charge will experience a force. It's created by charged objects, and its strength and direction are determined by the charge distribution. In the case of parallel plates with opposite charges, the electric field is relatively uniform, meaning it has the same strength and direction at all points between the plates (except near the edges, which we'll ignore for now).

The strength of the electric field is measured in Newtons per Coulomb (N/C), which tells us the force that would be exerted on a 1 Coulomb charge placed in the field. The direction of the electric field is defined as the direction of the force that would be exerted on a positive charge. So, in our case, the electric field points from the positive plate towards the negative plate.

Next, we have the concept of electric force. When a charged particle is placed in an electric field, it experiences a force. The magnitude of this force is given by the product of the charge of the particle and the strength of the electric field: F = qE, where F is the force, q is the charge, and E is the electric field strength. The direction of the force depends on the sign of the charge. A positive charge will experience a force in the same direction as the electric field, while a negative charge will experience a force in the opposite direction.

In our problem, the charged particle is positive, so it will experience a force pushing it away from the positive plate and towards the negative plate. This force will cause the particle to accelerate, meaning its velocity will change over time. To figure out exactly how the particle will move, we'll need to use Newton's laws of motion.

Finally, let's not forget about the concept of electric potential difference, often called voltage. The electric potential difference between two points is the amount of work required to move a unit positive charge from one point to the other. In the case of parallel plates, the electric potential difference is related to the electric field strength and the distance between the plates: V = Ed, where V is the potential difference, E is the electric field strength, and d is the distance. This concept will be useful for calculating the energy gained by the particle as it moves between the plates.

The Question: What's Next for the Charge?

The core question we need to answer is: what happens to the charged particle once it's placed on the positive plate? Does it stay there? Does it move? If it moves, how fast does it go, and how long does it take to reach the negative plate? To answer these questions, we need to use our knowledge of electric fields, forces, and motion.

Firstly, we know that the charged particle is positive and the plate it's sitting on is also positive. Like charges repel, so the particle will experience an electric force pushing it away from the positive plate. This force is what will set the particle in motion. Now, let's calculate the magnitude of this force using the formula F = qE. We have q = 2 x 10^-6 C and E = 6 x 10^7 N/C, so:

F = (2 x 10^-6 C) * (6 x 10^7 N/C) = 120 N

That's a pretty significant force for such a tiny particle! This force is the driving force behind the particle's acceleration. To find the acceleration, we'll use Newton's second law of motion: F = ma, where F is the force, m is the mass, and a is the acceleration. We know F = 120 N and m = 5 x 10^-6 kg, so we can solve for a:

a = F / m = 120 N / (5 x 10^-6 kg) = 2.4 x 10^7 m/s^2

Wow! The particle is accelerating at an incredible rate! This means its velocity is increasing rapidly as it moves towards the negative plate.

Calculating the Motion: Kinematics to the Rescue!

Now that we know the acceleration, we can use kinematic equations to figure out how long it takes the particle to reach the negative plate and how fast it's going when it gets there. Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. We have a constant acceleration, so we can use the following kinematic equation:

d = v₀t + (1/2)at^2

where d is the distance, v₀ is the initial velocity, t is the time, and a is the acceleration. In our case, the particle starts from rest, so v₀ = 0. The distance d is the distance between the plates, which is 0.1 meters. We already know a = 2.4 x 10^7 m/s^2. Plugging these values into the equation, we get:

0. 1 m = 0 + (1/2)(2.4 x 10^7 m/s^2)t2

Solving for t, we get:

t = √(2 * 0.1 m / (2.4 x 10^7 m/s^2)) ≈ 2.89 x 10^-5 s

So, it takes the particle about 28.9 microseconds (that's 28.9 millionths of a second!) to reach the negative plate. That's incredibly fast!

Now, let's find the final velocity of the particle when it reaches the negative plate. We can use another kinematic equation:

v = v₀ + at

We know v₀ = 0, a = 2.4 x 10^7 m/s^2, and t ≈ 2.89 x 10^-5 s. Plugging these values in, we get:

v = 0 + (2.4 x 10^7 m/s^2)(2.89 x 10^-5 s) ≈ 694 m/s

The particle is traveling at approximately 694 meters per second when it hits the negative plate! That's faster than the speed of sound!

Energy Considerations: Potential and Kinetic

Another way to think about this problem is in terms of energy. The charged particle initially has electric potential energy due to its position in the electric field. As it moves towards the negative plate, this potential energy is converted into kinetic energy, the energy of motion.

The electric potential difference between the plates is V = Ed = (6 x 10^7 N/C)(0.1 m) = 6 x 10^6 V (Volts). The electric potential energy of the particle on the positive plate is U = qV = (2 x 10^-6 C)(6 x 10^6 V) = 12 J (Joules). This is the amount of energy the particle has available to convert into kinetic energy.

As the particle reaches the negative plate, all of its potential energy has been converted into kinetic energy. The kinetic energy is given by KE = (1/2)mv^2. We can use this to check our previous calculation of the final velocity:

12 J = (1/2)(5 x 10^-6 kg)v^2

Solving for v, we get:

v = √(2 * 12 J / (5 x 10^-6 kg)) ≈ 693 m/s

This is very close to our previous result, which confirms that our calculations are consistent.

Key Takeaways

So, guys, we've seen how a charged particle behaves in a uniform electric field created by parallel plates. We calculated the electric force on the particle, its acceleration, the time it takes to travel between the plates, and its final velocity. We also looked at the energy transformation from potential energy to kinetic energy.

This problem illustrates some fundamental concepts in electromagnetism and mechanics. Understanding how electric fields affect charged particles is crucial for many applications, from particle accelerators to electronic devices. Hopefully, this deep dive has given you a solid understanding of the physics involved!

Further Exploration

Want to explore this topic further? You could investigate how the motion of the particle changes if the electric field is not uniform, or if there's a magnetic field present as well. You could also consider what happens if the particle is not a point charge, but an extended object with its own charge distribution. The possibilities are endless! Keep exploring, keep questioning, and keep learning!