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Chapter 23: Problem 35

A very small sphere with positive charge \(q=+8.00 \mu \mathrm{C}\) is released from rest at a point \(1.50 \mathrm{~cm}\) from a very long line of uniform linear charge density \(\lambda=+3.00 \mu \mathrm{C} / \mathrm{m} .\) What is the kinetic energy of the sphere when it is \(4.50 \mathrm{~cm}\) from the line of charge if the only force on it is the force exerted by the line of charge?

Short Answer

Expert verified
The kinetic energy of the sphere when it is \(4.50\, \mathrm{cm}\) from the line of charge is equal to the work done on the sphere by the electric field, calculated using the values provided in the problem.

Step by step solution

01

Calculating the Electric Field

The magnitude of the electric field \(E\) at some point a distance \(r\) perpendicular to an infinite line of charge is given by Coulomb鈥檚 law: \(E = \frac{{k\lambda}}{{r}}\), where \(k = \frac{1}{{4\pi\epsilon_0}} = 9 \times 10^9\, \mathrm{N \cdot m^2/C^2}\) is Coulomb's constant, and \(\epsilon_0\, \mathrm{C^2 / N \cdot m^2}\) is the permittivity of free space. Then the electric field at \(1.50\, \mathrm{cm}\) and \(4.50\, \mathrm{cm}\) from the long line charge can be calculated as following: for location 1, \(E_1 = \frac{{k\lambda}}{{r_1}}\), and for location 2, \(E_2 = \frac{{k\lambda}}{{r_2}}\).
02

Calculating the Force

The electric force \(F\) on a charge in an electric field is given by \(F = qE\), where \(q\) is the charge and \(E\) is the electric field. At location 1 and 2, the forces can be calculated as follows: at location 1, \(F_1 = qE_1\), and at location 2, \(F_2 = qE_2\). Both forces are directed along the line connecting the sphere and the line of charge, and are calculated knowing the values of \(q, E_1, E_2\).
03

Calculating the work done

The work \(W\) done on the sphere by the electric field as the sphere moves from location 1 to location 2 is given by \(W = F_1d_1 - F_2d_2\), where \(d_1\) and \(d_2\) are the distances of the sphere from the charged line at locations 1 and 2 respectively. \(W\) can be calculated by substituting the calculated values of \(F_1, F_2, d_1, d_2\) into the equation.
04

Applying the work-energy theorem

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. As the sphere starts from rest, its initial kinetic energy is zero. Therefore, the final kinetic energy \(K.E.\) of the sphere when it is \(4.50\, \mathrm{cm}\) from the line of charge is equal to the work done on the sphere by the electric field, i.e., \(K.E. = W\).

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle in electromagnetism, named after the French physicist Charles-Augustin de Coulomb. It provides a way to calculate the electric force between two point charges. According to Coulomb's Law, the force (\( F \)) between two charges is directly proportional to the product of their charges (\( q_1 \times q_2 \) ), and inversely proportional to the square of the distance (\( r^2 \) ) between them.

The equation is represented as \( F = k \times \frac{q_1 \times q_2}{r^2} \) , where \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \) ), illustrating the strength of the electrostatic interaction. In the context of our problem, Coulomb's Law helps to derive the formula for the electric field (\( E \) ) created by a linear charge density (\( \lambda \) ).

This foundational concept is critical for understanding electric fields, forces, and ultimately, how charges move in response to these forces, which directly connects to the behavior of the charged sphere in our textbook problem.
Electric Field
The electric field is a vector field around a charged object, representing the force that other charges would experience if placed within this field. The strength of the electric field is defined by the force felt per unit charge. Its units are newtons per coulomb (\( \mathrm{N/C} \) ).

For an infinite line of charge, the electric field (\( E \) ) at a distance \( r \) from the line of charge is described by \( E = \frac{{k\lambda}}{{r}} \) , where \( \lambda \) is the linear charge density and \( k \) is Coulomb's constant. Thus, the electric field diminishes as the distance from the line of charge increases. This directly informs our calculation of forces exerted on the charged sphere and ultimately its kinetic energy.
Work-Energy Theorem
The work-energy theorem is a critical concept linking the work done by forces to changes in kinetic energy. It states that the net work done by all forces acting on an object will result in an equivalent change in the object鈥檚 kinetic energy. Mathematically, it's expressed as \( W = \Delta KE \) , where \( W \) is the work and \( \Delta KE \) is the change in kinetic energy.

In the case of our charged sphere, as it accelerates from rest under the effect of the electric field, all the work done on the sphere by the electric force converts into its kinetic energy. This connection allows us to use the calculated work to find the final kinetic energy of the sphere once it has moved a certain distance away from the line of charge.
Linear Charge Density
Linear charge density (\( \lambda \) ) is essentially a measure of electric charge per unit length. In equations, it's denoted as \( \lambda = \frac{Q}{L} \) , where \( Q \) is the total charge, and \( L \) is the length over which the charge is distributed.

For continuous charge distributions, such as the long line of charge in our problem, it is a valuable concept to quantify the charge distribution. Knowing the linear charge density, \( \lambda \) , allows us to calculate the electric field, and subsequently the force that acts on the charged sphere. The uniformity of \( \lambda \) greatly simplifies the force and energy calculations that are essential to solving this problem regarding the kinetic energy of the sphere.

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Most popular questions from this chapter

Two stationary point charges \(+3.00 \mathrm{nC}\) and \(+2.00 \mathrm{nC}\) are separated by a distance of \(50.0 \mathrm{~cm}\). An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is \(10.0 \mathrm{~cm}\) from the \(+3.00 \mathrm{nC}\) charge?

A heart cell can be modeled as a cylindrical shell that is \(100 \mu \mathrm{m}\) long, with an outer diameter of \(20.0 \mu \mathrm{m}\) and a cell wall thickness of \(1.00 \mu \mathrm{m}\), as shown in Fig. \(\mathrm{P} 23.81 .\) Potassium ions move across the cell wall, depositing positive charge on the outer surface and leaving a net negative charge on the inner surface. During the so-called resting phase, the inside of the cell has a potential that is \(90.0 \mathrm{mV}\) lower than the potential on its outer surface. (a) If the net charge of the cell is zero, what is the magnitude of the total charge on either cell wall membrane? Ignore edge effects and treat the cell as a very long cylinder. (b) What is the magnitude of the electric field just inside the cell wall? (c) In a subsequent depolarization event, sodium ions move through channels in the cell wall, so that the inner membrane becomes positively charged. At the cnd of this event, the inside of the cell has a potential that is \(20.0 \mathrm{mV}\) higher than the potential outside the cell. If we model this event by charge moving from the outer membrane to the inner membrane, what magnitude of charge moves across the cell wall during this event? (d) If this were done entirely by the motion of sodium ions, \(\mathrm{Na}^{+}\), how many ions have moved?

A gold nucleus has a radius of \(7.3 \times 10^{-15} \mathrm{~m}\) and a charge of \(+79 e\). Through what voltage must an alpha particle, with charge \(+2 e,\) be accelerated so that it has just enough energy to reach a distance of \(2.0 \times 10^{-14} \mathrm{~m}\) from the surface of a gold nucleus? (Assume that the gold nucleus remains stationary and can be treated as a point charge.)

(a) How much excess charge must be placed on a copper sphere \(25.0 \mathrm{~cm}\) in diameter so that the potential of its center is \(3.75 \mathrm{kV} ?\) Take the point where \(V=0\) to be infinitely far from the sphere, (b) What is the potential of the sphere's surface?

A metal sphere with radius \(r_{a}\) is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius \(\eta_{b}\). There is charge \(+q\) on the inner sphere and charge \(-q\) on the outer spherical shell. (a) Calculate the potential \(V(r)\) for (i) \(rn\). (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take \(V\) to be zero when \(r\) is infinite. (b) Show that the potential of the inner sphere with respect to the outer is $$ V_{a b}=\frac{q}{4 \pi \epsilon_{0}}\left(\frac{1}{r_{a}}-\frac{1}{n_{b}}\right) $$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude \(E(r)=\frac{V_{a b}}{\left(1 / r_{a}-1 / r_{b}\right)} \frac{1}{r^{2}}\) (d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance \(r\) from the center, where \(r>r_{b} .\) (e) Suppose the charge on the outer sphere is not \(-q\) but a negative charge of different magnitude, say \(-Q .\) Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different
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