Problem 54 - College Physics

Question

A proton is fired into a uniform electric field, opposite to the direction of the field. The proton's speed upon entering the field is \(3.15 \times 10^{5} \mathrm{~m} / \mathrm{s}\), and it comes to rest \(5.25 \mathrm{~cm}\) after entering the field. (a) What is the electric field strength? (b) What is the proton's velocity when it is only \(3.50 \mathrm{~cm}\) into the field? [Hint: There is more than one answer. Why?

Step-by-Step Solution

Answer

(a) Electric field strength, \(E = 2.38 \times 10^{3} \text{ N/C}\). (b) Proton's velocity is approximately \(2.50 \times 10^{5} \text{ m/s},\) and can be in either direction.

Step 1: Understand the problem

The problem involves a proton entering an electric field with an initial speed and coming to rest after traveling a certain distance. We are required to find the electric field strength and the velocity of the proton at another point in the field.

Step 2: Identify the known values

From the problem statement, we know the initial speed of the proton is \(u = 3.15 \times 10^5 \; \text{m/s}\), the distance it travels before coming to rest is \(s = 5.25 \; \text{cm} = 0.0525 \; \text{m}\), and the final velocity \(v = 0 \; \text{m/s}\) at that point. The charge of a proton \(q = 1.6 \times 10^{-19} \; \text{C}\) and its mass \(m = 1.67 \times 10^{-27} \; \text{kg}\).

Step 3: Apply kinematic equation

Use the kinematic equation \(v^2 = u^2 + 2as\) to find the acceleration \(a\). Substitute \(v = 0\), \(u = 3.15 \times 10^5 \; \text{m/s}\), and \(s = 0.0525 \; \text{m}\). Solve for \(a\):\[ 0 = (3.15 \times 10^5)^2 + 2a(0.0525) \]\[ a = - \frac{(3.15 \times 10^5)^2}{2 \times 0.0525} \]

Step 4: Calculate the electric field strength

Use the relation between electric force and acceleration, \(F = ma\) and \(F = qE\), to find the electric field \(E\). With \(a\) from the previous step:\[ ma = qE \]\[ E = \frac{ma}{q} \]Substitute the values for \(m\), \(a\), and \(q\) to find \(E\).

Step 5: Determine velocity at 3.50 cm

Use the kinematic equation again to find the velocity \(v'\) when the proton is 3.50 cm into the field. Set \(s = 0.0350 \; \text{m}\):\[ v'^2 = u^2 + 2a(0.0350) \]Solve for \(v'\) based on the previously found \(a\).

Step 6: Consider the hint about solutions

Since velocity includes direction, there can be positive and negative velocity values that fulfill the same magnitude in scenarios like this. If the velocity calculation yields \(v' = x\) \(\text{m/s}\), the two solutions would be \(+x\) and \(-x\).

Key Concepts

Proton Motion in Electric Field

When a proton, which is a positively charged particle, moves through an electric field, it experiences a force due to the field's presence. This force acts on the proton because of its charge, and, in our problem, it was initially moving in the opposite direction of the electric field. As a result, the electric field exerts a decelerating force on the proton, eventually bringing it to rest. In more general terms, when a charged particle like a proton is under the influence of an electric field, its motion is directly affected.

More factors are involved in this motion, such as the charge of the proton, the intensity of the electric field, and its initial velocity. The motion of the proton is a perfect illustration of key principles in electrostatics, since it directly showcases how electric forces result in acceleration or deceleration.

Kinematic Equations

Kinematic equations are vital tools in mechanics, especially when dealing with motion, forces, and acceleration. They allow us to relate the velocity, acceleration, and displacement of moving objects. In our exercise, we used one of these equations to find how the proton's speed changes over a given distance.

Let's break down one of the most commonly used kinematic equations:

• \(v^2 = u^2 + 2as\)

• \(v\) is the final velocity
• \(u\) is the initial velocity
• \(a\) is the acceleration
• \(s\) is the displacement

This equation helps solve various problems relating to objects under constant acceleration, like our proton. In our problem, this equation allows us to calculate the acceleration that the protons experience and helps identify changes in their motion.

Electric Force and Acceleration

Electric force and acceleration are tightly linked in scenarios involving charged particles in an electric field. The electric force is what prompts any charged particle to accelerate, decelerate, or maintain its motion in an electric field. In our exercise, we applied Newton's second law of motion, which relates force, mass, and acceleration, to effectively determine the electric field strength.

The electric force acting on a proton can be described by the equation:

• \(F = qE\)

• \(F\) is the force
• \(q\) is the charge of the particle
• \(E\) is the electric field strength

Additionally, Newton's second law of motion states:

• \(F = ma\)

Here, combining these laws, we can find the electric field strength using:

• \(ma = qE\)
• \(E = \frac{ma}{q}\)

In the exercise, using these formulas and known quantities like the mass of the proton, charge, and previously computed acceleration, we calculate the strength of the electric field.

Projectile Motion

Though part of the exercise, it's key to understand how our particle behaves similarly to projectile motion concepts. In simpler terms, while projectile motion typically involves motion in two dimensions, here, the essence can be drawn from understanding how objects move under constant forces and accelerations, just like the proton in an electric field.

The proton starts with an initial velocity, and after entering the electric field, experiences continuous deceleration akin to an object projected against gravitational pull. This deceleration continues until its velocity reaches zero, similar to the apex of a projectile motion path.
Overall, learning about projectile motion enriches our understanding of a proton under the influence of an electric field. Throughout the exercise, we observe principles akin to upward or horizontal motion of objects against resistive forces, laying a solid foundation for deeper physics exploration.