/*! 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 67 Self-Energy of a Sphere of Charg... [FREE SOLUTION] | 91Ó°ÊÓ

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Self-Energy of a Sphere of Charge. A solid sphere of radius \(R\) contains a total charge \(Q\) distributed uniformly throughout its volume. Find the energy needed to assemble this charge by bringing infinitesimal charges from far away. This encrgy is called the "self-cnergy" of the charge distribution. (Hint: After you have assembled a charge \(q\) in a sphere of radius \(r\), how much energy would it take to add a spherical shell of thickness dr having charge dq? Then integrate to get the total energy.)

Short Answer

Expert verified
The self-energy of the solid sphere with uniformly distributed charge is \(\frac{3kQ^2}{5R}\).

Step by step solution

01

Understand the Concept

The self-energy in a charge distribution is the work done to assemble the charge distribution from infinitesimally small charges each brought from far away (infinity).
02

Set Up the Integral for Energy

Considering a small spherical shell of thickness \(dr\) at a distance \(r\) from the center of the sphere and it carries a charge amount \(dq\), the small work \(dU\) needed to place this charge is given by the formula \(dU = \frac{k \cdot dq \cdot q}{r}\), where \(dq = \tfrac{3Q}{4\pi R^3}4\pi r^2 dr\). Substitute this into the formula for \(dU.\)
03

Carry Out the Integration

We compute the total energy \(U\) by integrating \(dU\) from \(0\) to \(Q\) for \(q\) and from \(0\) to \(R\) for \(r\). This will give the total work done: \(U = \frac{3kQ^2}{5R}\).

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Key Concepts

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

Electrostatics
Electrostatics is a branch of physics that studies electric charges at rest. It encompasses the behavior of charges, the forces they exert on each other, and the electric fields they create. Understanding electrostatics is crucial for solving problems related to charge distribution, such as the self-energy of a sphere of charge. Self-energy is defined as the energy needed to assemble a charge distribution by bringing infinitesimal charges from infinity to their final location.

When calculating the self-energy of a charge distribution, we consider the work done against the electrostatic forces. These forces result from the interactions between charges, described by Coulomb's Law. Coulomb's Law states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Charge Distribution
Charge distribution refers to how electric charge is arranged within a material or object. In the given exercise, we have a solid sphere with a uniform charge distribution. The charge distribution is mathematically expressed in terms of charge density, which can be volume, surface, or linear density, depending on the object’s shape. For a sphere with uniform volume charge density, we assume the charge is spread evenly throughout the entire volume.

In this exercise, the uniform charge density allows us to calculate the charge on an infinitesimal element of the sphere, such as a thin spherical shell of thickness dr, using the formula dq = (3Q/4Ï€R3)4Ï€°ù2dr. This approach simplifies the integration process when determining the total self-energy, which involves summing up the energy required to assemble each charge element.
Integration in Physics
Integration in physics is a mathematical tool used for adding up quantities that are distributed over a certain range, such as charge or energy in the problem discussed. To find the self-energy of the charge distribution, we set up an integral that sums the incremental work done to bring each charge element from infinity.

Integration allows us to calculate properties for the whole charge distribution from knowing how properties behave for infinitesimal parts of the distribution. In this case, we integrate the work dU = (k dq q) / r over the entire volume of the sphere. Intuitively, it's like calculating the total amount of paint needed for a sculpture by summing up the paint required for each tiny area of the surface.
Gaussian Surface
A Gaussian surface is a hypothetical closed surface used in Gauss's Law, a key concept in electrostatics. Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. This law is immensely useful in calculating electric fields of symmetrical charge distributions.

While our current problem does not directly involve the application of Gauss's Law, understanding Gaussian surfaces is still beneficial. It provides insight into electric field behavior and aids in appreciating how the charge within a given volume contributes to the electrostatic potential at every point in space. This background knowledge supports the idea of calculating work done, or potential energy, in a charge distribution.
Electric Potential Energy
Electric potential energy is the energy that a charged object possesses due to its position in an electric field. When calculating the self-energy of a sphere, we're essentially finding the electric potential energy required to assemble the sphere from a dispersed state to a uniform sphere of radius R.

The work needed to move a charge dq to a sphere that already has charge q is proportional to q and inversely proportional to the distance r from the center of the sphere. Integrating this work from zero up to the total charge Q gives us the sphere's self-energy, which is the potential energy stored in the charge distribution due to the electrostatic forces between the charges.

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Most popular questions from this chapter

Identical charges \(q=+5.00 \mu \mathrm{C}\) are placed at opposite corners of a square that has sides of length \(8.00 \mathrm{~cm}\). Point \(A\) is at one of the empty corners, and point \(B\) is at the center of the square. A charge \(q_{0}=-3.00 \mu \mathrm{C}\) is placed at point \(A\) and moves along the diagonal of the square to point \(B\). (a) What is the magnitude of the net electric force on \(\bar{q}_{0}\) when it is at point \(A ?\) Sketch the placement of the charges and the direction of the net force. (b) What is the magnitude of the net electric force on \(q_{0}\) when it is at point \(B ?\) (c) How much work does the electric force do on \(q_{0}\) during its motion from \(A\) to \(B ?\) Is this work positive or negative? When it goes from \(A\) to \(B,\) does \(q_{0}\) move to higher potential or to lower potential?

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?

A solid conducting sphere of radius \(5.00 \mathrm{~cm}\) carries a net charge. To find the value of the charge, you measure the potential difference \(V_{A B}=V_{A}-V_{B}\) between point \(A,\) which is \(8.00 \mathrm{~cm}\) from the center of the sphere, and point \(B\), which is a distance \(r\) from the center of the sphere. You repeat these measurements for several values of \(r>8.00 \mathrm{~cm} .\) When you plot your data as \(V_{A B}\) versus \(1 / r,\) the values lie close to a straight line with slope \(-18.0 \mathrm{~V} \cdot \mathrm{m}\). What does your data give for the net charge on the sphere? Is the net charge positive or negative?

An alpha particle with kinetic energy \(9.50 \mathrm{MeV}\) (when far away) collides head-on with a lead nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and may be treated as a point charge. The atomic number of lead is \(82 .\) The alpha particle is a helium nucleus, with atomic number \(2 .\) )

A small sphere with mass \(5.00 \times 10^{-7} \mathrm{~kg}\) and charge \(+7.00 \mu \mathrm{C}\) is released from rest a distance of \(0.400 \mathrm{~m}\) above a large horizontal insulating sheet of charge that has uniform surface charge density \(\sigma=+8.00 \mathrm{pC} / \mathrm{m}^{2} .\) Using energy methods, calculate the speed of the sphere when it is \(0.100 \mathrm{~m}\) above the sheet.

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