/*! 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 34 A fixed circle of radius a and c... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 34

A fixed circle of radius a and centre \(O\) is drawn and a charge \(+q\) is placed at a distance \(3 a / 4\) from \(O\) on a line through \(O\) perpendicular to the plane of circle. If a second charge \(q^{\prime}\) is similarly placed at a distance \(5 a / 12\) on the opposite side of circle, the net electric flux through circle becomes zero. Find the ratio of \(q^{\prime}\) to \(q\).

Short Answer

Expert verified
The ratio of \( q' \) to \( q \) is \( \frac{13}{40} \).

Step by step solution

01

Understanding the problem

We are given a circle with its center at point \(O\) and two charges placed on a line perpendicular to the plane of the circle. The solution involves calculating the electric flux through the circle due to both charges and ensuring their net flux sums to zero.
02

Electric flux from a point charge

The electric flux \( \Phi \) due to a point charge \( q \) at a distance \( d \) from the center of a circle (when the circle lies in a plane perpendicular to the line connecting the charge to the center) is given by \( \Phi = \frac{q}{2 \varepsilon_0} \left[ 1 - \frac{d}{\sqrt{d^2 + a^2}} \right] \), where \( a \) is the radius of the circle and \( \varepsilon_0 \) is the permittivity of free space.
03

Calculate flux for first charge

Insert the known values for the first charge \(+q\) into the flux formula: \( \Phi_1 = \frac{q}{2 \varepsilon_0} \left[ 1 - \frac{3a/4}{\sqrt{(3a/4)^2 + a^2}} \right] \). Simplifying the expression in the square root \( \sqrt{(9a^2/16) + a^2} = \sqrt{25a^2/16} = (5/4)a \).
04

Calculate flux for second charge

Insert the known values for the second charge \(q'\): \( \Phi_2 = \frac{q'}{2 \varepsilon_0} \left[ 1 - \frac{5a/12}{\sqrt{(5a/12)^2 + a^2}} \right] \). Simplify the square root \( \sqrt{25a^2/144 + a^2} = \sqrt{169a^2/144} = (13/12)a \).
05

Set net flux to zero

Since the net flux must sum to zero, set \( \Phi_1 + \Phi_2 = 0 \): \[ \frac{q}{2 \varepsilon_0} \left[ 1 - \frac{3/4}{5/4} \right] + \frac{q'}{2 \varepsilon_0} \left[ 1 - \frac{5/12}{13/12} \right] = 0. \]
06

Solve for \(q'/q\)

Solve the equation \( \Phi_1 + \Phi_2 = 0 \) to find \( \frac{q'}{q} \). Simplifying, we have: \[ \frac{q}{2 \varepsilon_0} \left( \frac{1}{5} \right) = \frac{q'}{2 \varepsilon_0} \left( \frac{8}{13} \right). \] Therefore, \( \frac{q'}{q} = \frac{13}{40}. \)

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Key Concepts

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

Point Charge
A point charge is an idealized model of a charged particle. It's assumed to have a negligible size, allowing us to treat the charge as if it were concentrated at a single point in space. This simplification is useful for calculating electric fields and electric flux because it eliminates the complexity of charge distribution over an area.
A positive point charge, like the charge denoted as \(+q\) in the given exercise, contributes to the electric flux through a surface. The direction of its electric field is radially outward, following Coulomb's law. The strength of the field decreases with the square of the distance from the charge, which affects the amount of electric flux passing through surrounding surfaces.
  • Single Concentrated Point: The charge is conceptualized as a single point.
  • Radial Electric Field: The field lines emanate outwards in all directions.
  • Simplified Calculations: An ideal model for calculating electric fields in theoretical problems.
Understanding the nature of point charges helps in grasping how they interact with surfaces to create an electric flux through those surfaces.
Permittivity of Free Space
Permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental constant that describes how much electric field is permitted to "flow" through the vacuum of space. It is a measure of the inherent resistance that space offers to the electric field. The value of \( \, \varepsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \), where F/m stands for farads per meter, which measures capacitance.
This constant appears in many electric equations, such as Coulomb鈥檚 law and the expression for calculating electric flux. In the context of the exercise, it forms part of the denominator in the electric flux equation, affecting how much flux the charge \( q \) generates through the circle.
  • Fundamental Constant: Appears in equations about electric fields in a vacuum.
  • Measurement Unit: Measured in farads per meter.
  • Influences Electric Flux: Affects the calculations of electric fields generated by charges.
For students studying electromagnetism, understanding permittivity of free space is crucial to predict how electric fields behave in vacuum scenarios.
Electric Flux Calculation
Electric flux quantifies the number of electric field lines passing through a given area. In mathematical terms, when a point charge is involved, electric flux \( \Phi \) is described by:\[ \Phi = \frac{q}{2 \varepsilon_0} \left[ 1 - \frac{d}{\sqrt{d^2 + a^2}} \right] \]where \( q \) is the charge, \( d \) is the distance from the charge to the center of the circle, and \( a \) is the circle's radius.
This equation is used to find how much flux is caused by a point charge through a surface like the circle in the exercise. The formula stems from integrating the electric field over the surface area while considering the geometry of how this field intersects the surface.
  • Integration: Derived from integrating the electric field over the surface area.
  • Dependence on Geometry: Considers the perpendicular distance and the radius of the circle.
  • Balance of Fluxes: Requires calculating for each charge and setting their sum to zero for total neutrality.
By carefully applying this formula, students can determine how charges affect a surface, which was crucial in finding the ratio \( \frac{q'}{q} \) in the original exercise.

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Most popular questions from this chapter

A region in space contains a total positive charge \(Q\) that is distributed spherically such that the volume charge density \(\rho(r)\) is given by $$ \begin{array}{ll} \rho(r)=3 \alpha r /(2 R) & \text { for } r \leq R / 2 \\ \rho(r)=\alpha\left[1-(r / R)^{2}\right] & \text { for } R / 2 \leq r \leq R \\\ \rho(r)=0 & \text { for } r \geq R \end{array} $$ Here \(\alpha\) is a positive constant having units of \(\mathrm{C} / \mathrm{m}^{3} .\) (a) Determine \(\alpha\) in terms of \(Q\) and \(R\). (b) Using Gauss's law, derive an expression for the magnitude of the electric field as a function of \(r .\) Do this separately for all three regions. Express your answers in terms of the total charge \(Q\). (c) What fraction of the total charge is contained within the region \(R / 2 \leq r \leq R ?\) (d) What is the magnitude of \(\vec{E}\) at \(r=R / 2 ?\) (e) If an electron with charge \(q^{\prime}=-e\) is released from rest at any point in any of the three regions, the resulting motion will be oscillatory but not simple harmonic. Why?

Negative charge \(-Q\) is distributed uniformly over the surface of a thin spherical insulating shell with radius \(R\). Calculate the force (magnitude and direction) that the shell exerts on a positive point charge \(q\) located (a) a distance \(r>R\) from the center of the shell (outside the shell) and (b) a distance \(r

A negative charge \(-Q\) is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. (a) Is there any excess charge induced on the inner surface of the piece of metal? If so, find its sign and magnitude. (b) Is there any excess charge on the outside of the piece of metal? Why or why not? (c) Is there an electric field in the cavity? Explain. (d) Is there an electric field within the metal? Why or why not? Is there an electric field outside the piece of metal? Explain why or why not. (e) Would someone outside the solid measure an electric field due to the charge \(-Q ?\) Is it reasonable to say that the grounded conductor has shielded the region from the effects of the charge \(-Q\) ? In principle, could the same thing be done for gravity? Why or why not?

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter \(12.0 \mathrm{~cm}\), giving it a charge of \(-36 \mu \mathrm{C}\). Find the electric field (a) just inside the paint layer; (b) just outside the paint layer; (c) \(6 \mathrm{~cm}\) outside the surface of the paint layer.

A point charge of \(+4.9 \mu \mathrm{C}\) is located on the \(x\)-axis at \(x=4.00 \mathrm{~m}\), next to a spherical surface of radius \(3.00 \mathrm{~m}\) centered at the origin. (a) Calculate the magnitude of the electric field at \(x=3.00 \mathrm{~m} .\) (b) Calculate the magnitude of the electric field at \(x=-3.00 \mathrm{~m}\). (c) According to Gauss's law, the net flux through the sphere is zero because it contains no charge. Yet the field due to the extemal charge is much stronger on the near side of the sphere (i.e., at \(x=3.00 \mathrm{~m}\) ) than on the far side (at \(x=-3.00 \mathrm{~m}\) ). How, then, can the flux into the sphere (on the near side) equal the flux out of it (on the far side)? Explain. A sketch will help.
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