/*! 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 31 Charge \(Q=5.00 \mu \mathrm{C}\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Charge \(Q=5.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=12.0 \mathrm{~cm} .\) A small sphere with charge \(q=+3.00 \mu \mathrm{C}\) and mass \(6.00 \times 10^{-5} \mathrm{~kg}\) is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within \(8.00 \mathrm{~cm}\) of the surface of the large sphere?

Short Answer

Expert verified
The minimum speed \(v\) that the small sphere must have in order to come within \(8.00 \mathrm{~cm}\) of the surface of the large sphere can be determined by substituting the problem specifics into the equation \(v = \sqrt{\frac{2U}{m}}\).

Step by step solution

01

Calculate the Electric Field of the large sphere

The electric field inside a uniformly charged sphere at a distance \(r\) from the center is given by \(E = k \frac{Qr}{R}\) where \(Q\) is the total charge, \(R\) is radius of the sphere, and \(k\) is Coulomb's constant \((9.0 \times 10^9 N m^2/C^2)\). Here, \(r = R - 8cm = 0.04m\) because we want the electric field 8 cm within the surface of the large sphere.
02

Compute the potential energy at the near point

The potential energy of the small sphere near the surface of the large sphere can be calculated by the formula \(U = k\frac{Qq}{r}\). The total energy of the small sphere is conserved, so the kinetic energy at the large distance (when potential energy is zero), is equal to the potential energy at \(r = 0.04m\). Hence, \(U = \frac{1}{2}mv^2\) where \(m\) is mass of the small sphere and \(v\) is its speed.
03

Solve for the minimum speed

We rearrange the equation from step 2 to solve for \(v\), giving us \(v = \sqrt{\frac{2U}{m}} = \sqrt{\frac{2kQq}{mr}}\). Using the data provided in the problem, we can substitute and solve for \(v\).

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.

Electric Field
Imagine a magical force that surrounds positive or negative charged objects, communicating with every other charged object around them. This is what we call an electric field. It is a region where an electric charge, if placed, would experience a force. We measure electric fields in units of volts per meter (V/m). They can attract or repel charged objects depending on the nature of the charge—like charges repel, unlike charges attract.

The electric field inside a uniformly charged sphere at any point can be calculated with a simple formula: \( E = k \frac{Qr}{R^3} \).
  • Here, \(E\) stands for electric field, \(k\) is Coulomb's constant, \(Q\) is the total charge, \(r\) is the distance from the center, and \(R\) is the radius of the sphere.
The formula tells us that within the sphere, the electric field increases as you move away from the center toward the surface. Outside the sphere, it changes based on different conditions and is calculated differently, but that's a story for another day! Since our specific problem is just concerned with electric fields inside the sphere, this formula suffices.

In the exercise, we are particularly interested in the electric field a short distance inside the sphere's surface. This is essential information when calculating potential energy, which we'll explain next.
Potential Energy
Potential energy in the realm of electrostatics is like stored energy that can perform work. In the case of charged particles, this energy is due to the positions of the charges relative to one another. It’s crucial because as the small sphere comes closer to the charged sphere, the electric field exerts a force on it, which affects the speed it needs to have.

Potential energy \( U \) between two charges can be calculated by \( U = k\frac{Qq}{r} \), where
  • \(Q\) and \(q\) are the charges of the large and small sphere,
  • \(r\) is the distance between them,
  • \(k\) remains Coulomb’s constant, telling us the strength of the force per unit charge.
The potential energy signifies how much work it would take to bring the small charge from infinity (a big distance away) to a certain point closer to the large sphere.

In the context of the solution, this energy at the point where the small sphere comes very close to the large sphere equals the kinetic energy it had initially set off with. Understanding this swap between potential and kinetic energy is where conservation of energy joins our story.
Conservation of Energy
Conservation of Energy is a fundamental principle in physics stating that in a closed system, the total energy remains constant. This means energy can change forms (from kinetic to potential, and vice versa), but the total energy doesn't change. For our exercise, this principle allows us to compare the energy of the small sphere at two different points.

Initially, at a large distance from the big sphere, the small sphere has maximum kinetic energy, calculated with \( \frac{1}{2}mv^2 \), where:
  • \(m\) is the mass of the sphere,
  • \(v\) is its speed,
  • and the potential energy is nearly zero because it's so far away.
When it becomes close enough to feel the big sphere's electric field, its kinetic energy transforms into potential energy. The exciting part here is when the small sphere slows down due to electrostatic forces as it encounters increased potential energy.

Thus, understanding the conservation of energy allows you to calculate the minimum speed the small sphere needs to avoid falling short of its target distance from the large sphere. It bridges what happens to the energy as it hurdles into and within the charged field. Such principles provide profound insights into how charged bodies interact and move within electric fields.

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

A thin insulating rod is bent into a semicircular arc of radius \(a,\) and a total electric charge \(Q\) is distributed uniformly along the rod. Calculate the potential at the center of curvature of the are if the potential is assumed to be zero at infinity.

CP Deflecting Plates of an Oscilloscope. The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying cqual but opposite charges. Typical dimensions are about \(3.0 \mathrm{~cm}\) on a side, with a separation of about \(5.0 \mathrm{~mm} .\) The potential difference between the plates is \(25.0 \mathrm{~V}\). The plates are close enough that we can ignore fringing at the cnds. Under these conditions: (a) how much charge is on each plate, and (b) how strong is the electric field between the plates? (c) If an electron is ejected at rest from the negative plate, how fast is it moving when it reaches the positive plate?

An infinitely long line of charge has linear charge density \(5.00 \times 10^{-12} \mathrm{C} / \mathrm{m} .\) A proton (mass \(1.67 \times 10^{-27} \mathrm{~kg},\) charge \(+1.60 \times 10^{-19} \mathrm{C}\) ) is \(18.0 \mathrm{~cm}\) from the line and moving directly toward the line at \(3.50 \times 10^{3} \mathrm{~m} / \mathrm{s}\). (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge?

A point charge \(q_{1}=+5.00 \mu \mathrm{C}\) is held fixed in space. From a horizontal distance of \(6.00 \mathrm{~cm},\) a small sphere with mass \(4.00 \times 10^{-3} \mathrm{~kg}\) and charge \(q_{2}=+2.00 \mu \mathrm{C}\) is fired toward the fixed charge with an initial speed of \(40.0 \mathrm{~m} / \mathrm{s}\). Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is \(25.0 \mathrm{~m} / \mathrm{s} ?\)

For a particular experiment, helium ions are to be given a kinetic energy of \(3.0 \mathrm{MeV}\). What should the voltage at the center of the accelerator be, assuming that the ions start essentially at rest? (a) \(-3.0 \mathrm{MV}\) (b) \(+3.0 \mathrm{MV} ;(\mathrm{c})+1.5 \mathrm{MV} ;(\mathrm{d})+1.0 \mathrm{MV}\)

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.