Problem 38 - University Physics with Modern Physics

Question

CP A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 \(\mathrm{cm}\) distant from the first, in a time interval of \(1.50 \times 10^{-6}\) s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

Step-by-Step Solution

Answer

(a) The magnitude of the electric field is approximately 1487.37 N/C. (b) The speed of the proton when it strikes the plate is about 21300 m/s.

Step 1: Analyze the Situation

We have two parallel plates with a uniform electric field \(E\) between them. A proton moves from the positively charged plate to the negatively charged plate covering a distance \(d = 1.60\text{ cm} = 0.016\text{ m}\) in a time \(t = 1.50 \times 10^{-6}\text{ s}\). The proton is initially at rest.

Step 2: Use Kinematics to Find Acceleration

The proton starts from rest, so initial velocity \(u = 0\). The distance \(d\) it travels in time \(t\) can be expressed as \(d = \frac{1}{2} a t^2\). Solving for acceleration \(a\), we have: \[ a = \frac{2d}{t^2} = \frac{2 \times 0.016\text{ m}}{(1.50 \times 10^{-6}\text{ s})^2} \approx 1.42 \times 10^{10} \text{ m/s}^2 \]

Step 3: Calculate the Magnitude of the Electric Field

The force on the proton is given by \( F = ma \) where \( m = 1.67 \times 10^{-27} \text{ kg} \) is the mass of the proton. In an electric field, the force is also \( F = qE \), where \( q = 1.60 \times 10^{-19} \text{ C} \) is the charge of the proton. Setting \( ma = qE \), we solve for \(E\): \[ E = \frac{ma}{q} = \frac{1.67 \times 10^{-27} \text{ kg} \times 1.42 \times 10^{10} \text{ m/s}^2}{1.60 \times 10^{-19} \text{ C}} \approx 1487.37 \text{ N/C} \]

Step 4: Use Final Velocity Formula to Find Speed

The final velocity \(v\) can be found using \(v = u + at\). Since the proton starts from rest \(u = 0\), it simplifies to \(v = at\): \[ v = (1.42 \times 10^{10} \text{ m/s}^2) \times (1.50 \times 10^{-6} \text{ s}) \approx 21300 \text{ m/s} \]

Key Concepts

Proton motion

Protons are positively charged particles found in the nucleus of an atom. Understanding their motion is crucial, especially in an electric field, because it provides insights into how charged particles behave under the influence of forces. Protons have a specific charge and mass: the charge is approximately \( 1.60 \times 10^{-19} \text{ C} \), and the mass is about \( 1.67 \times 10^{-27} \text{ kg} \). When a proton is released in an electric field, the field's force acts on it. This force is determined by the equation \( F = qE \), where \( q \) is the charge of the proton and \( E \) is the electric field. The acceleration of the proton, therefore, depends on this force and the mass of the proton because \( F = ma \). By knowing these fundamentals, we can predict a proton's path once it enters an electric field. In the context of the homework problem, this motion is from a positively charged plate to a negatively charged plate.

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the causes of this motion, such as forces or energy. It allows us to describe motion through concepts like velocity, acceleration, and displacement. By using kinematic equations, we can solve for quantities such as distance traveled, velocity at any point, and time of travel. In this problem, we use the equation \( d = \frac{1}{2} a t^2 \) to find the proton's acceleration because it starts at rest and moves a known distance in a specified time. Breaking it down:

• \( d \) is the displacement of the proton.
• \( a \) is the acceleration we're solving for.
• \( t \) is the time the proton takes to reach the other plate.

Once we have acceleration from this equation, we can find the proton's final velocity using \( v = u + at \), where \( u = 0 \) due to the initial rest condition.

Uniform electric field

A uniform electric field is a region where electric force is constant in magnitude and direction. Imagine pulling a rope with a steady hand; the tension felt throughout is similar to how a uniform electric field behaves. In this exercise, the region between the charged plates is considered to have such a field. This means that anywhere between the plates, the electric force on a charge is the same. This uniformity simplifies calculations significantly.
For the proton, this field applies an unchanging force throughout its journey from one plate to another. To find the strength or magnitude of the electric field, we use the relationship \( F = qE \), with \( E = \frac{F}{q} \). Here, knowing the proton's acceleration and mass lets us find the force involved. This field helps us predict not only acceleration but also ensure the proton travels a known path.

Charged plates

Charged plates create an electric field in the space between them, similar to the magnetic field created between two magnets. In this problem, these plates are oppositely charged, creating a straightforward path for the proton to follow. The charged plates setup is often used in physics to demonstrate electric fields because it gives a visual and simple understanding of how fields operate. When a charge, such as a proton, is placed in this field, it will feel a force and move accordingly from the positively charged plate towards the negatively charged one. The distance between the plates and the uniformly charged condition are crucial, as they help dictate the field's intensity. For any exercises involving electric fields like this, remembering that particles will always move towards the opposite charge is key. This setup not only illustrates electric field lines but also showcases the predictability of charged particle motion.