/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 LA A proton is placed in a unifo... [FREE SOLUTION] | 91Ó°ÊÓ

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LA 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\(\mu\) in the field, assuming it starts from rest.

Short Answer

Expert verified
Force: \(4.40 \times 10^{-16}\,\text{N}\), Acceleration: \(2.63 \times 10^{11}\,\text{m/s}^2\), Speed: \(263\,\text{m/s}\).

Step by step solution

01

Calculate the Electric Force

The force exerted on a charged particle in an electric field can be calculated using the formula \( F = qE \), where \( q \) is the charge of the proton \( (1.60 \times 10^{-19} \text{ C}) \) and \( E \) is the electric field \( (2.75 \times 10^3 \text{ N/C}) \).\[ F = (1.60 \times 10^{-19} \, \text{C}) \times (2.75 \times 10^3 \, \text{N/C}) \]Calculate this product to find the force felt by the proton.
02

Calculate the Proton's Acceleration

To find the acceleration, use Newton's second law, \( F = ma \), where \( m \) is the mass of a proton \( (1.67 \times 10^{-27} \, \text{kg}) \). Rearrange the formula to solve for acceleration \( a \): \( a = \frac{F}{m} \).Substitute the value of the force from Step 1 and the mass of the proton:\[ a = \frac{4.40 \times 10^{-16} \, \text{N}}{1.67 \times 10^{-27} \, \text{kg}} \]Perform the division to find the acceleration of the proton.
03

Calculate the Proton's Speed After 1.00 Microsecond

The speed can be calculated using the formula \( v = u + at \), where \( u \) is the initial velocity (0 m/s), \( a \) is the acceleration found in Step 2, and \( t \) is the time \( (1.00 \, \mu\text{s} = 1.00 \times 10^{-6} \, \text{s}) \).\[ v = 0 + a \times 1.00 \times 10^{-6} \, \text{s} \]Substitute the value of acceleration and calculate \( v \).
04

Conclusion

The magnitude of the electric force is \( 4.40 \times 10^{-16} \, \text{N} \), the acceleration is \( 2.63 \times 10^{11} \, \text{m/s}^2 \), and the speed of the proton after \( 1.00 \, \mu\text{s} \) is \( 263 \, \text{m/s} \).

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Key Concepts

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

Proton Acceleration
When we talk about proton acceleration in an electric field, we're diving into a fundamental concept of physics. The acceleration of a charged particle, like a proton, is directly influenced by the forces acting on it. In this scenario, the force is due to the electric field.

Acceleration can be determined using Newton's second law, which states that force equals mass times acceleration: \( F = ma \). By rearranging the equation, we can solve for acceleration \( a \) as follows:
  • \( a = \frac{F}{m} \)
Where \( F \) is the force exerted on the proton, and \( m \) is the mass of the proton. This tells us how fast the proton's velocity changes as it moves under the influence of the electric field. Remember, the greater the force, or the smaller the mass, the higher the acceleration.

In our specific exercise, once we calculate the force using the proton's charge and the electric field strength, we can find out precisely how quickly the proton will accelerate.
Electric Field
An electric field represents how an electrical charge exerts force on other charges around it. It's like an invisible "force field" that spreads out from charge sources.

In a uniform electric field, like the one in our exercise, the electric field lines are parallel and evenly spaced, which means that the force experienced by a charge within it is constant. The formula \( F = qE \) is employed to find the force on a charge:
  • \( F \) is the force on the charge.
  • \( q \) represents the charge of the proton, which is positive.
  • \( E \) is the electric field strength given in Newtons per Coulomb (\( \text{N/C} \)).
This interaction is why a proton, when placed in an electric field, experiences a force that pushes it in the direction of the field.

Understanding electric fields helps us predict and control how charges behave, which is fundamental not only in academic settings but also in real-world applications like designing electronic circuits and even in understanding natural phenomena.
Fundamental Physics Equations
Physics is full of equations that help us understand and predict the universe's behavior. In this problem, we focus on a few key equations that simplify these processes.

The main equation here, \( F = qE \), allows us to calculate the force on a charge due to an electric field. It brings together the concepts of force, charge, and field strength.

Newton's second law, \( F = ma \), is another cornerstone. It connects force, mass, and acceleration, highlighting the direct relationship between these quantities.

Finally, the kinematic equation \( v = u + at \) lets us determine the speed of an object after a specific period, given constant acceleration. This formula helps calculate how the proton's speed will evolve over time, starting from rest and under constant acceleration.
  • \( v \) is the final velocity.
  • \( u \) is the initial velocity.
  • \( a \) is the acceleration.
  • \( t \) is the time interval.
Understanding these equations gives students the ability to solve a wide array of problems, not just those involving protons but in various domains of physics.

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Most popular questions from this chapter

Point charge \(q_{1}=-5.00 \mathrm{nC}\) is at the origin and point charge \(q_{2}=+3.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=3.00 \mathrm{cm} .\) Point \(P\) is on the \(y\) -axis at \(y=4.00 \mathrm{cm} .\) (a) Calculate the electric fields \(\vec{\boldsymbol{E}}_{1}\) and \(\vec{\boldsymbol{E}}_{2}\) at point \(P\) due to the charges \(q_{1}\) and \(q_{2} .\) Express your results in terms of unit vectors (see Example 21.6 ). (b) Use the results of part (a) to obtain the resultant field at \(P\) , expressed in unit vector form.

Two very large parallel sheets are 5.00 \(\mathrm{cm}\) apart. Sheet \(A\) carries a uniform surface charge density of \(-9.50 \mu \mathrm{C} / \mathrm{m}^{2},\) and sheet \(B,\) which is to the right of \(A,\) carries a uniform charge density of \(-11.6 \mu \mathrm{C} / \mathrm{m}^{2}\) . Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 \(\mathrm{cm}\) to the right of sheet \(A\) (b) 4.00 \(\mathrm{cm}\) to the left of sheet \(A ;(c) 4.00 \mathrm{cm}\) to the right of sheet \(B\) .

\(\mathrm{A}-5.00\) -nC point charge is on the \(x\) -axis at \(x=1.20 \mathrm{m.}\) A second point charge \(Q\) is on the \(x\) -axis at \(-0.600 \mathrm{m} .\) What must be the sign and magnitude of \(Q\) for the resultant electric field at the origin to be (a) 45.0 \(\mathrm{N} / \mathrm{C}\) in the \(+x\) -x-direction, (b) 45.0 \(\mathrm{N} / \mathrm{C}\) in the - \(x\) -direction?

BIO Estimate how many electrons there are in your body. Make any assumptions you feel are necessary, but clearly state what they are. (Hint: Most of the atoms in your body have equal numbers of electrons, protons, and neutrons.) What is the combined charge of all these electrons?

A dipole consisting of charges \(\pm e, 220 \mathrm{nm}\) apart, is placed between two very large (essentially infinite) sheets carrying equal but opposite charge densities of 125\(\mu \mathrm{C} / \mathrm{m}^{2}\) . (a) What is the maximum potential energy this dipole can have due to the sheets, and how should it be oriented relative to the sheets to attain this value? (b) What is the maximum torque the sheets can exert on the dipole, and how should it be oriented relative to the sheets to attain this value? (c) What net force do the two sheets exert on the dipole?

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