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A proton is placed in a uniform electric field of \(2.75 \times\) \(10^{3} \mathrm{N} / \mathrm{C}\) . Calculate: (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after 1.00\(\mu \mathrm{s}\) in the field, assuming it starts from rest.

Short Answer

Expert verified
(a) Force: \(4.40 \times 10^{-16} \text{ N}\); (b) Acceleration: \(2.63 \times 10^{11} \text{ m/s}^2\); (c) Speed: \(2.63 \times 10^{5} \text{ m/s}\).

Step by step solution

01

Calculate the Electric Force on the Proton

To find the electric force experienced by the proton, use the formula \( F = qE \), where \( F \) is the force, \( q \) is the charge of the proton, and \( E \) is the electric field. The charge of a proton is \( q = 1.60 \times 10^{-19} \, \text{C} \). Thus, the force is:\[F = (1.60 \times 10^{-19} \, \text{C})(2.75 \times 10^{3} \, \text{N/C}) = 4.40 \times 10^{-16} \, \text{N}\]
02

Calculate the Proton's Acceleration

The acceleration \( a \) can be found using Newton's second law \( F = ma \), where \( m \) is the mass of the proton \((m = 1.67 \times 10^{-27} \, \text{kg})\). Rearrange to find:\[a = \frac{F}{m} = \frac{4.40 \times 10^{-16} \, \text{N}}{1.67 \times 10^{-27} \, \text{kg}} = 2.63 \times 10^{11} \, \text{m/s}^2\]
03

Determine the Proton's Speed After 1.00 \(\mu\text{s}\)

Use the formula for final velocity for constant acceleration, \( v = u + at \), where \( u \) is the initial velocity (0 m/s since it starts from rest), \( a \) is the acceleration, and \( t \) is the time (\(1.00 \, \mu\text{s} = 1.00 \times 10^{-6} \, \text{s}\)). So:\[v = 0 + (2.63 \times 10^{11} \, \text{m/s}^2)(1.00 \times 10^{-6} \, \text{s}) = 2.63 \times 10^{5} \, \text{m/s}\]

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Key Concepts

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

Electric Force
The electric force is a fundamental interaction that occurs between charged particles. It describes how these particles exert influence over each other within an electric field. The strength of this force depends directly on the charge of the particles and the electric field strength.
The formula for calculating the electric force (\( F \)) on a charge (\( q \)) in an electric field (\( E \)) is given by:
  • \( F = qE \)
For a proton, which has a charge of \( q = 1.60 \times 10^{-19} \, \text{C} \), the electric force in an electric field with a magnitude of \( 2.75 \times 10^{3} \, \text{N/C} \) can be calculated as:
  • \( F = (1.60 \times 10^{-19} \, \text{C})(2.75 \times 10^{3} \, \text{N/C}) = 4.40 \times 10^{-16} \, \text{N} \)
This calculation shows how a relatively small charge, like that of a proton, can experience significant forces due to a strong electric field.
Proton Acceleration
When a proton is subjected to an electric force, it accelerates. The acceleration of the proton can be determined using Newton's second law, which states that the force on an object is equal to its mass times its acceleration.
The formula for finding acceleration (\( a \)) is:
  • \( F = ma \)
  • \( a = \frac{F}{m} \)
The mass of a proton is \( m = 1.67 \times 10^{-27} \, \text{kg} \). Using the previously calculated force of \( 4.40 \times 10^{-16} \, \text{N} \), the acceleration can be calculated as:
  • \( a = \frac{4.40 \times 10^{-16} \, \text{N}}{1.67 \times 10^{-27} \, \text{kg}} = 2.63 \times 10^{11} \, \text{m/s}^2 \)
This result indicates that protons can achieve very high accelerations when exposed to strong electric fields.
Uniform Electric Field Effect
A uniform electric field implies that the field strength is consistent across a specified region. This means that any charged particle within this field will experience a constant force and therefore undergo constant acceleration.
In this context, a proton placed in a uniform electric field will have predictable motion. It begins to accelerate immediately due to the electric force acting on it, and this force does not change because the field is uniform. The knowledge of such consistent acceleration allows us to determine parameters like speed and kinetic energy over time, as demonstrated in the specific application of this uniform electric field to a proton.
  • Uniform fields simplify calculations as the force \( F \) and acceleration \( a \) remain constant.
  • This consistency allows for straightforward application of kinematic equations to find velocity and displacement at any given time.
Newton's Second Law
Newton's Second Law of Motion is a fundamental principle that describes how the velocity of an object changes when it is subjected to an external force. It is expressed through the equation:
  • \( F = ma \)
This law articulates that the force exerted on an object is equal to the mass of the object multiplied by its acceleration.
In the case of a proton in an electric field, Newton’s Second Law helps us understand how the electric force (\( F \)) influences the proton's acceleration (\( a \)). Given that the mass (\( m \)) of the proton is a known value, we can rearrange the formula to solve for the acceleration:
  • \( a = \frac{F}{m} \)
This approach provides an insightful way to determine the change in motion of a proton when it is acted upon by an electric force. The principle is central to solving many physics problems involving forces and motion.

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Most popular questions from this chapter

An infinite sheet with positive charge per unit area \(\sigma\) lies in the \(x y\) -plane. A second infinite sheet with negative charge per unit area \(-\sigma\) lies in the \(y z\) -plane. Find the net electric field at all points that do not lie in either of these planes. Express your answer in terms of the unit vectors \(\hat{i}, \hat{j},\) and \(\hat{k} .\)

Sketch the electric field lines for a disk of radius \(R\) with a positive uniform surface charge density \(\sigma\) . Use what you know about the electric field very close to the disk and very far from the disk to make your sketch.

Strength of the Electric Force. Imagine two \(1.0-\mathrm{g}\) bags of protons, one at the earth's north pole and the other at the south pole. (a) How many protons are in each bag? (b) Calculate the gravitational attraction and the electrical repulsion that each bag exerts on the other. (c) Are the forces in part (b) large enough for you to feel if you were holding one of the bags?

A proton is traveling horizontally to the right at \(4.50 \times\) \(10^{6} \mathrm{m} / \mathrm{s}\) (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of \(3.20 \mathrm{cm} .\) (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

A negative charge \(-0.550 \mu C\) exerts an upward \(0.200-\mathrm{N}\) force on an unknown charge 0.300 \(\mathrm{m}\) directly below it. (a) What is the unknown charge (magnitude and sign)? (b) What are the magnitude and direction of the force that the unknown charge exerts on the \(-0.550-\mu \mathrm{C}\) charge?

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