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Chapter 23: Problem 44

A proton is projected in the positive \(x\) direction into a region of a uniform electric field \(\mathbf{E}=-6.00 \times 10^{5} \hat{\mathbf{i}} \mathrm{N} / \mathrm{C}\) at \(t=0 .\) The proton travels 7.00 \(\mathrm{cm}\) before coming to rest. Determine (a) the acceleration of the proton, (b) its initial speed, and (c) the time at which the proton comes to rest.

Short Answer

Expert verified
The acceleration (a) of the proton is \(3.60 \times 10^{13} \text{ m/s}^2\). The initial speed (vi) is \(1.80 \times 10^{6} \text{ m/s}\). The time (t) to come to rest is \(5.00 \times 10^{-8} \text{ s}\).

Step by step solution

01

Calculate the Acceleration

Use Newton's second law which states that the net force on an object is equal to the product of its mass and its acceleration, \(F = ma\). The force experienced by a proton in an electric field is given by \(F = qE\), where \(q\) is the charge of the proton and \(E\) is the electric field. Since the electric field is uniform, we can set these two expressions equal to solve for acceleration, \(-6.00 \times 10^{5} \times q = ma\). The charge of a proton is approximately \(1.60 \times 10^{-19} \text{C}\) and its mass is roughly \(1.67 \times 10^{-27} \text{kg}\). Plugging in these values, we solve for \(a\).
02

Determine the Initial Speed

We use the kinematic equation \(v_{f}^{2} = v_{i}^{2} + 2a(x_{f} - x_{i})\), where \(v_{f}\) is the final velocity, \(v_{i}\) the initial velocity, \(a\) the acceleration, and \(x_{f} - x_{i}\) the displacement. Since the proton comes to rest, \(v_{f} = 0\). The displacement is given as 7.00 cm which we convert to meters (0.07 m). Plugging in known values and solving for \(v_{i}\) gives us the initial speed.
03

Calculate the Time to Come to Rest

We can use the kinematic equation \(v_{f} = v_{i} + at\), where \(t\) is the time. Since \(v_{f} = 0\), we can solve for \(t\) by rearranging the equation to \(t = -\frac{v_{i}}{a}\). We use the initial speed from Step 2 and the acceleration from Step 1 to find the time it takes for the proton to come to rest.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Understanding Newton's second law is critical when analyzing the motion of particles under the influence of forces. This fundamental principle of physics can be summarized as follows: the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In mathematical terms, this is expressed as \( F = ma \), where \( F \) represents the net force, \( m \) the mass, and \( a \) the acceleration.

In our exercise with the proton in an electric field, we apply this law to determine the electric force acting on the proton. Since the force is the product of the charge of the proton (\( q \) = 1.60 脳 10^-19 C) and the electric field (\( E \) = -6.00 脳 10^5 N/C), we can then equate this to \( ma \) to find the proton's acceleration. Understanding this relationship helps in predicting the proton's behavior under the known electric field.
Electric Force on Charged Particles
When dealing with electric fields, it's crucial to understand how they exert force on charged particles. The force \( F \) acting on a charged particle with charge \( q \) in an electric field \( E \) can be calculated using the simple equation \( F = qE \). This is a vector equation, meaning both the force and the electric field have magnitude and direction.

In our case, the electric field is given in the negative \( x \) direction, meaning it will exert a force on the proton that is opposite to its initial direction of motion. This force is what causes the proton to eventually stop. Thus, analyzing the direction and magnitude of the electric field provides us with valuable insights into the path and final state of the proton.
Kinematic Equations
The kinematic equations are a set of formulas that describe the motion of objects in classical mechanics. These equations relate variables like velocity, acceleration, time, and displacement without considering the forces that cause such motion. They are extremely useful for solving problems involving uniform acceleration, like our proton's travel through an electric field.

One such equation is \( v_{f}^{2} = v_{i}^{2} + 2a(x_{f} - x_{i}) \), which we used to find the proton's initial speed, knowing its final speed (zero, as it comes to rest), acceleration (from Newton's second law), and displacement. Another is \( v_{f} = v_{i} + at \), which allowed us to solve for the time it took for the proton to stop. By mastering these equations, students can predict an object's future state of motion given its current state and the accelerations acting upon it.

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Most popular questions from this chapter

A proton accelerates from rest in a uniform electric field of 640 \(\mathrm{N} / \mathrm{C}\) . At some later time, its speed is \(1.20 \times 10^{6} \mathrm{m} / \mathrm{s}\) (nonrelativistic, because \(v\) is much less than the speed of light). (a) Find the acceleration of the proton. (b) How long does it take the proton to reach this speed? (c) How far has it moved in this time? (d) What is its kinetic energy at this time?

Two point charges are located on the \(x\) axis. The first is a charge \(+Q\) at \(x=-a\) . The second is an unknown charge located at \(x=+3 a\) . The net electric field these charges produce at the origin has a magnitude of 2\(k_{e} Q / a^{2}\) . What are the two possible values of the unknown charge?

A uniformly charged ring of radius 10.0 \(\mathrm{cm}\) has a total charge of 75.0\(\mu \mathrm{C}\) . Find the electric field on the axis of the ring at (a) \(1.00 \mathrm{cm},\) (b) \(5.00 \mathrm{cm},(\mathrm{c}) 30.0 \mathrm{cm},\) and (d) 100 \(\mathrm{cm}\) from the center of the ring.

Consider \(n\) equal positive point charges each of magnitude \(Q / n\) placed symmetrically around a circle of radius \(R .\) (a) Calculate the magnitude of the electric field at a point a distance \(x\) on the line passing through the center of the circle and perpendicular to the plane of the circle. (b) Explain why this result is identical to that of the calculation done in Example 23.8 .

Two identical conducting small spheres are placed with their centers 0.300 \(\mathrm{m}\) apart. One is given a charge of 12.0 \(\mathrm{nC}\) and the other a charge of \(-18.0 \mathrm{nC}\) . (a) Find the electric force exerted by one sphere on the other. (b) What If? The spheres are connected by a conducting wire. Find the electric force between the two after they have come to equilibrium.
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