/*! 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 12 12\. A particle is uncharged and... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

12\. A particle is uncharged and is thrown vertically upward from ground level with a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) . As a result, it attains a maximum height \(h\) . The particle is then given a positive charge \(+q\) and reaches the same maximum height \(h\) when thrown vertically upward with a speed of 30.0 \(\mathrm{m} / \mathrm{s}\) . The electric potential at the height \(h\) exceeds the electric potential at ground level. Finally, the particle is given a negative charge -q Ignoring air resistance, determine the speed with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the maximum height h. In all three situations, be sure to include the effect of gravity.

Short Answer

Expert verified
The initial speed for the negatively charged particle should be approximately 6.12 m/s.

Step by step solution

01

Understand the Problem

We're dealing with a particle thrown upwards under the force of gravity and electric potential. Initially, the particle is uncharged, then positively charged, and finally negatively charged. We need to find the initial speed for the negatively charged scenario such that it reaches the same height. The difference in behavior between charged and uncharged conditions arises from electric potential differences.
02

Analyze the First Situation (Uncharged Particle)

For the uncharged particle, we apply energy conservation. Initially, it has kinetic energy and no potential energy. At maximum height, all energy is potential:\[ \frac{1}{2} m v_1^2 = mgh \]where \(v_1 = 25.0\; \text{m/s}\) and \(h\) is the maximum height. Simplify to find \(gh\).
03

Analyze the Second Situation (Positively Charged Particle)

For the positively charged particle, the mechanical energy equation includes electric potential energy. The equation becomes:\[ \frac{1}{2} m v_2^2 = mgh + qV \]where \(v_2 = 30.0\; \text{m/s}\). We've already calculated \(mgh\) from the first scenario, so solve \(qV = \frac{1}{2}m(v_2^2 - 25^2)\).
04

Find the Charge-Related Energy Difference

From Step 3, we state that:\[ qV = \frac{1}{2}m(30^2 - 25^2) \]Handling these values:\[ qV = \frac{1}{2}m \times 275 \]
05

Analyze the Third Situation (Negatively Charged Particle)

For the negatively charged particle, energy at the maximum involves negative potential energy contribution:\[ \frac{1}{2} m v_3^2 + qV = mgh \]Here, \(v_3\) is the unknown speed, and rearranging gives:\[ v_3^2 = gh - \frac{2qV}{m} \].
06

Simplify and Solve for \(v_3\)

Using relationships from Steps 2 to 5 with \(qV = \frac{1}{2}m \times 275\) and \(gh = 312.5\), solve:\[ v_3^2 = 312.5 - 275 \]\[ v_3 = \sqrt{37.5} \]
07

Calculate the Final Answer

Calculate the value:\[ v_3 = \sqrt{37.5} \approx 6.12\; \text{m/s} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In physics, it is defined as half of the mass of the object multiplied by the square of its velocity. The general formula is:
  • \( K = \frac{1}{2} m v^2 \)
where \( K \) is the kinetic energy, \( m \) is the mass, and \( v \) is the velocity of the object. In the given problem, the initial kinetic energy of the particle plays a critical role in determining how high it will go when launched. For an uncharged particle, its kinetic energy is completely converted into gravitational potential energy at its maximum height. For the charged scenarios, the kinetic energy has to overcome additional electric potential energy alongside gravitational potential energy. This means, with different initial velocities for the positively and negatively charged particles, how the particle converts kinetic energy into total potential energy varies.
Electric Charge
Electric charge refers to a fundamental property of particles that cause them to experience a force when placed in an electromagnetic field. Charges can be positive or negative, and they influence electric potential energy, which is the work needed to move a charge against an electric field.
In the context of the exercise, the moving particle experiences different effects based on its charge:
  • When positively charged, it experiences an extra push due to the positive electric potential at height \( h \).
  • When negatively charged, it experiences an opposing pull in the same environment.
The electric charge modifies how much kinetic energy the particle needs to reach the maximum height. A positively charged particle needs less initial kinetic energy, because it gains electric potential energy, whereas a negatively charged particle needs more to overcome the additional potential energy barrier.
Conservation of Energy
The principle of conservation of energy states that in an isolated system, energy cannot be created or destroyed, only transformed from one form to another. In mechanical systems, total energy is the sum of kinetic and potential energies.
When the particle in this exercise is thrown upward:
  • Initial kinetic energy is gradually converted into gravitational potential energy as it ascends.
  • For charged particles, electric potential energy also comes into play. It can either aid or oppose the motion depending on the charge.
In each situation — uncharged, positively charged, and negatively charged — calculations utilize the conservation of energy to deduce the initial speed needed to reach the height. This principle helps us understand that the energy adjustments correspond to changes in electric and gravitational interactions. By analyzing energy transformations, we can precisely solve for the particle's velocity under different conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two identical point charges \(\left(q=+7.20 \times 10^{-6} \mathrm{C}\right)\) are fixed at diagonally opposite corners of a square with sides of length 0.480 \(\mathrm{m} .\) A test charge \(\left(q_{0}=-2.40 \times 10^{-8} \mathrm{C}\right)\) with a mass of \(6.60 \times 10^{-8} \mathrm{kg},\) is released from rest at one of the empty corners of the square. Determine the speed of the test charge when it reaches the center of the square.

Multiple-Concept Example 4 to see the concepts that are pertinent here. In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 25000 \(\mathrm{V}\) . The speeds of the electrons are quite large, and for accurate calculations of the speeds, the effects of special relativity must be taken into account. Ignoring such effects, find the electron speed just before the electron strikes the screen.

A capacitor is constructed of two concentric conducting cylindrical shells. The radius of the inner cylindrical shell is \(2.35 \times 10^{-3} \mathrm{m},\) and that of the outer shell is 2.50 \(\times 10^{-3} \mathrm{m}\) . When the cylinders carry equal and opposite charges of magnitude 1.7 \(\times 10^{-10} \mathrm{C}\) , the electric field between the plates has an average magnitude of \(4.2 \times 10^{4} \mathrm{V} / \mathrm{m}\) and is directed radially outward from the inner shell to the outer shell. Determine (a) the magnitude of the potential difference between the cylindrical shells and \((\text { b) the capacitance of this capacitor. }\)

Suppose that the electric potential outside a living cell is higher than that inside the cell by 0.070 V. How much work is done by the electric force when a sodium ion (charge \(=+e )\) moves from the outside to the inside?

Two identical capacitors store different amounts of energy: capacitor A stores \(3.1 \times 10^{-3} \mathrm{J}\) and capacitor \(\mathrm{B}\) stores \(3.4 \times 10^{-4} \mathrm{J}\) . The voltage across the plates of capacitor \(\mathrm{B}\) is 12 \(\mathrm{V}\) . Find the voltage across the plates of capacitor A.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.