Physics
Electric Field of a Continous Charge Distribution
The electric field of a continuous charge distribution describes the force experienced by a test charge at any point in space due to the distribution of electric charges. It is calculated using the principle of superposition, where the contributions from individual charges are summed. This concept is fundamental in understanding the behavior of electrically charged objects and the interactions between them.
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10 Key excerpts on "Electric Field of a Continous Charge Distribution"
- Raymond Serway, John Jewett(Authors)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
One way is to use the superposition principle Continuous Charge Distributions and Gauss’s Law 23 Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 616 Chapter 23 Continuous Charge Distributions and Gauss’s Law from Equation 22.10. The sum in that equation will become an integral over the distribution. The second means of finding the electric field for certain types of continuous distributions of charge is to use Gauss’s law. Gauss’s law is based on the inverse square behavior of the electric force between point charges. Although Gauss’s law is a direct consequence of Coulomb’s law, it is more convenient for calculating the electric fields of highly symmetric charge distributions and makes it possible to deal with complicated problems using qualitative reasoning. This new tool will be added to our kit of techniques to be used in evaluating electric fields and can be used in future chapters whenever we encounter the electric field due to a continuous, symmetric charge distribution. 23.1 Electric Field of a Continuous Charge Distribution In Chapter 22, we investigated the electric fields due to point charges and the effects of external electric fields on point charges. Equation 22.10 is useful for cal- culating the electric field due to a small number of charges. In many cases, we have a continuous distribution of charge rather than a collection of discrete charges. The charge in these situations can be described as continuously distributed along some line, over some surface, or throughout some volume.
eBook - ePub- A. L. Stanford, J. M. Tanner(Authors)
- 2014(Publication Date)
- Academic Press(Publisher)
12Calculation of Electric Fields
Publisher Summary
This chapter explains several techniques appropriate to the calculation of electric fields. Electric fields play a key role in electromagnetic physics. It is, therefore, important to be able to determine the electric fields associated with a wide variety of charge distributions. Certain symmetrical arrangements of charge produce electric fields that may be determined by a simple, powerful, problem-solving procedure. This process is based on a relationship called Gauss’s law, which, in turn, is based on the concept of electric flux. Electric fields and the way that electric fields are related to Gauss’s law permit the understanding of how static charges position themselves in and on conducting materials. A continuous charge distribution may be considered a collection of infinitesimal elements of charge, each of which causes an electric field at a field point located a distance from the charge element. Each element of charge within the continuous distribution of charge is treated as if it were a point charge causing an electric field. The summation of infinitesimal elements is, of course, an integration process.Electric fields play a key role in electromagnetic physics. It is, therefore, important that we be able to determine the electric fields associated with a wide variety of charge distributions. This chapter introduces several techniques appropriate to the calculation of electric fields that result from relatively simple distributions of charge.Our first consideration is the calculation of electric fields at points in space caused by collections of fixed point charges. This procedure will then be extended to include the calculation of electric fields caused by continuous charge distributions.Certain symmetrical arrangements of charge produce electric fields that may be determined by a simple, powerful, problem-solving procedure. This process is based on a relationship called Gauss’s law, which, in turn, is based on the concept of electric flux. Therefore, we will illustrate electric flux and see how it is used in Gauss’s law to calculate the electric fields of symmetrical charge distributions.
No longer available |Learn morePhysics for Scientists and Engineers
Foundations and Connections, Extended Version with Modern Physics
- Debora Katz(Author)
- 2016(Publication Date)
- Cengage Learning EMEA(Publisher)
The overall plan for finding the electric field due to a con- tinuous distribution of charged particles is based on the procedure we followed for a collection of charged parti- cles. The first step is to imagine slicing up the distribution into tiny pieces, with each piece so small that it can be modeled as a single charged particle. We consider just a small number of pieces—usually one or two—to get an expression for the electric field produced by those pieces. Then we integrate that expression to find the electric field produced by the entire continuous distribution. INTERPRET and ANTICIPATE As for a collection of charged particles, draw a diagram showing the electric field at some particular position. There are four elements to this diagram: 1. Labeled pieces of the source and an imaginary test particle. Imagine slicing up the distribution into small pieces. It is usually helpful to indicate two of these pieces on your sketch. In choosing which two pieces, exploit the symmetry of the problem. Also draw an imaginary test particle (assumed positive) at the location where you wish to find the electric field. If you are not told whether the charge distribution is positive or negative, assume it is positive. 2. The electric field vectors at the location of the test charge, as found from the electric force exerted on the test charge by each piece of the source you’ve indicated. The direction of the force is along the line Electric Field due to a Continuous Distribution of Charged Particles PROBLEM-SOLVING STRATEGY joining the test particle to the particular piece of the source. To get the length of the vector, estimate the magnitude of the force by taking into account the distance to the test particle. As long as the charge distribution is uniform, each piece has the same small amount of charge dq, which makes your esti- mation task easy.
eBook - PDF- Pierluigi Zotto, Sergio Lo Russo, Paolo Sartori(Authors)
- 2022(Publication Date)
- Società Editrice Esculapio(Publisher)
2.11 Field and Potential of Continuous Charge Distributions The direct evaluation of the electrostatic field generated by continuous charge distribu- tion is practically feasible only if a symmetry of the system, allowing to simplify the calcu- lations, is identified. Then an evaluation can be done only for few cases and for some of their variants. a) Uniformly Charged Indefinite Straight Wire A uniformly charged indefinite wire is a very long, rectilinear and unidimensional, meaning that it is so thin that its transverse size is negligible, object carrying charge depos- ited according to a constant linear charge density λ. The electrostatic field generated by a wire in a point P is given by the sum of the contributions d E of all the infini- tesimal charge elements dq distributed along the wire, that is E P = d E B ∫ = 1 4πε 0 dq r 2 u B ∫ = 1 4πε 0 λdz r 2 u wire ∫ , where axis z has been chosen directed along the wire upwards and u is a unit vector, whose direction is variable and defined by the line connecting the position of element dq and point P, oriented positive away from the wire. Consider a cylindrical coordinate reference system featuring its axis z parallel to the wire and whose origin is chosen on the point of the wire where the normal to the wire pass- ing by P crosses it. + + + + + + + + + + λ P dq dz u dE Electrostatic Field and Electrostatic Potential Chapter 2 26 The indefinite length of the wire implies that each charge element dq1 = λ dz, depos- ited on an infinitesimal length dz of the wire in position +z, has a symmetrical element dq2 = dq1 in position –z. Hence, the electro- static field is given by the sum of the contri- butions of pairs of identical charge elements lying in symmetrical positions with respect to axis z. Each pair of charge elements generates an electrostatic field d E = d E 1 + d E 2 = dE u r , because the dEz components of the field gen- erated by each element are equal and opposite.
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
Our goals in this chapter are to (1) define electric field, (2) discuss how to calculate it for various arrangements of charged particles and objects, and (3) dis- cuss how an electric field can affect a charged particle (as in making it move). The Electric Field A lot of different fields are used in science and engineering. For example, a tem- perature field for an auditorium is the distribution of temperatures we would find by measuring the temperature at many points within the auditorium. Similarly, we could define a pressure field in a swimming pool. Such fields are examples of scalar fields because temperature and pressure are scalar quantities, having only magnitudes and not directions. In contrast, an electric field is a vector field because it is responsible for conveying the information for a force, which involves both magnitude and direc- tion. This field consists of a distribution of electric field vectors E → , one for each point in the space around a charged object. In principle, we can define E → at some point near the charged object, such as point P in Fig. 22-2a, with this proce- dure: At P, we place a particle with a small positive charge q 0 , called a test charge because we use it to test the field. (We want the charge to be small so that it does not disturb the object’s charge distribution.) We then measure the electrostatic force F → that acts on the test charge. The electric field at that point is then E → = F → q 0 (electric field). (22-1) Because the test charge is positive, the two vectors in Eq. 22-1 are in the same direction, so the direction of E → is the direction we measure for F → . The magnitude of E → at point P is F/q 0 . As shown in Fig. 22-2b, we always represent an electric field with an arrow with its tail anchored on the point where the measurement is made.
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
Instead, particle 2 pushes by means of the electric field it has set up. Our goals in this chapter are to (1) define electric field, (2) discuss how to calculate it for various arrangements of charged particles and objects, and (3) dis- cuss how an electric field can affect a charged particle (as in making it move). The Electric Field A lot of different fields are used in science and engineering. For example, a tem- perature field for an auditorium is the distribution of temperatures we would find by measuring the temperature at many points within the auditorium. Similarly, we could define a pressure field in a swimming pool. Such fields are examples of scalar fields because temperature and pressure are scalar quantities, having only magnitudes and not directions. In contrast, an electric field is a vector field because it is responsible for conveying the information for a force, which involves both magnitude and direc- tion. This field consists of a distribution of electric field vectors E → , one for each point in the space around a charged object. In principle, we can define E → at some point near the charged object, such as point P in Fig. 22.1.2a, with this proce- dure: At P, we place a particle with a small positive charge q 0 , called a test charge because we use it to test the field. (We want the charge to be small so that it does not disturb the object’s charge distribution.) We then measure the electrostatic force F → that acts on the test charge. The electric field at that point is then E → = F → ___ q 0 (electric field). (22.1.1) Because the test charge is positive, the two vectors in Eq. 22.1.1 are in the same direction, so the direction of E → is the direction we measure for F → . The mag- nitude of E → at point P is F/q 0 . As shown in Fig. 22.1.2b, we always represent an electric field with an arrow with its tail anchored on the point where the measure- ment is made.
eBook - PDF- Y K Lim(Author)
- 1986(Publication Date)
- WSPC(Publisher)
Chapter I FUNDAMENTAL CONCEPTS AND EXPERIMENTAL LAWS Electrodynamics deals with the fields and radiation of moving charges. In describing the interaction between charges it is convenient, both mathematically and physically, to consider it, not as forces that act at a distance, but as the force exerted by the field set up by one charge on the other. This approach is in fact essential for charges in relative motion as electromagnetic effects are found to propagate with finite velocity. The four field vectors, Ej B, D and H, which are fundamental in Maxwell's electro-magnetic theory are introduced and discussed in this chapter in a phenomenologiaal manner. In addition, a short review is made of the experimental laws which lead to Maxwell's equations. 1.1 Electric Field Intensity E Electric field is said to exist at a point where a stationary particle experiences a force on account of its charge. The electric field intensity or electric field strength E is defined as the force per unit charge acting on a small positive charge q' introduced at that point. Let F be the electric force acting on the test charge, then by definition 2 . E - lim -L . (1.1) q' + OQ' The limit q' + Q is required in order that the introduction of the test charge will not significantly Influence the source; the field can then be described independently of the presence of a test charge. The finite magnitude of the elementary charge e does not permit the limiting process to be realized even in principle. The definition applies, therefore, to macroscopic phenomena only. For microscopic processes, the field is usually defined in terms of its source, assuming that the macroscopic laws governing the field-source relationship still apply. The simplest type of electric field is one that is set up by stationary charges, the electrostatic field. We shall confine ourselves in the first instance to free space.
eBook - PDF- Paul Peter Urone, Roger Hinrichs(Authors)
- 2012(Publication Date)
- Openstax(Publisher)
(a) The electric field is a vector quantity, with both parallel and perpendicular components. The parallel component ( E ∥ ) exerts a force ( F ∥ ) on the free charge q , which moves the charge until F ∥ = 0 . (b) The resulting field is perpendicular to the surface. The free charge has been brought to the conductor’s surface, leaving electrostatic forces in equilibrium. A conductor placed in an electric field will be polarized. Figure 18.31 shows the result of placing a neutral conductor in an originally uniform electric field. The field becomes stronger near the conductor but entirely disappears inside it. Figure 18.31 This illustration shows a spherical conductor in static equilibrium with an originally uniform electric field. Free charges move within the conductor, polarizing it, until the electric field lines are perpendicular to the surface. The field lines end on excess negative charge on one section of the surface and begin again on excess positive charge on the opposite side. No electric field exists inside the conductor, since free charges in the conductor would continue moving in response to any field until it was neutralized. Misconception Alert: Electric Field inside a Conductor Excess charges placed on a spherical conductor repel and move until they are evenly distributed, as shown in Figure 18.32. Excess charge is forced to the surface until the field inside the conductor is zero. Outside the conductor, the field is exactly the same as if the conductor were replaced by a point charge at its center equal to the excess charge. 712 Chapter 18 | Electric Charge and Electric Field This OpenStax book is available for free at http://cnx.org/content/col11406/1.9 Figure 18.32 The mutual repulsion of excess positive charges on a spherical conductor distributes them uniformly on its surface. The resulting electric field is perpendicular to the surface and zero inside.
- Andrei D. Polyanin, Alexei Chernoutsan(Authors)
- 2010(Publication Date)
- CRC Press(Publisher)
(P3. 2 . 1 . 1 ) This formula defines the strength of an electric field , E . It also answers the question what force the field exerts on any charge q , moving or stationary. In SI, the strength of an electric field is measured in N/C or V/m. For brevity, the term electric field is used to mean the strength of an electric field whenever this does not lead to confusion. ◮ Electrostatic field. Principle of superposition. A stationary electric field generated by a system of stationary charges is called an electrostatic field . The strength of an electrostatic field can be determined using Coulomb’s law (P3.1.2.2). 1. Electric field of a point charge. If a point charge q is placed in the field of a point charge Q , then it follows from (P3.1.2.2) and (P3.2.1.1) that E = k Q r 2 r r = E r r r , (P3. 2 . 1 . 2 ) where E r is the projection of E onto the radial direction. 2. Electric field of a system of point charges. Using the principle of superposition, we get E = E 1 + E 2 + · · · = summationdisplay i k Q i ( r – r i ) 2 r – r i | r – r i | . (P3. 2 . 1 . 3 ) For a continuously distributed charge, the summation must be replaced by integration (see (P3.1.1.1)). (Compare with the formulas for the gravitational field in Subsection P1.7.2.) Example 1. There is a charge continuously distributed along an arc of a circle of radius R with a linear density λ (Fig. P3.1). The arc angle is 2 α 0 . Find the strength of the electric field generated by the charge, at the circle center. Figure P3.1. Strength of the electric field generated by a charged arc of a circle. Solution. By symmetry, the field strength is directed along the y -axis, which passes through the midpoint of the arc and the circle center O . The strength of the field generated by the arc element dα at O equals dE = kdq/R 2 , where dq = λRdα . Projecting onto y and integrating yields E = E y = integraldisplay α 0 – α 0 k λRdα R 2 cos α = k 2 λ sin α 0 R .
- Ozgur Ergul(Author)
- 2021(Publication Date)
- Wiley(Publisher)
• It appears that the electric field created by a point electric charge is infinite at the location of the electric charge. We recall that the point electric charge itself is an idealization (a total amount of electric charge Q 0 squeezed into a single point), as discussed in Section 1.3.2 . Therefore, having an infinite electric field at the electric charge location is perfectly consistent with the source definition. z x y Q 3 Q n D n -. . . Q 2 Q 1 D 3 -D 2 -D 1 -Figure 1.42 Electric fields created by a set of static electric charges. The vector sum of these electric fields gives the total electric field at the observation point. Electric charges and field values are linearly related to each other. According to the super-position principle, linear effects can be added. For example, if there are n point electric charges Gauss’ law 45 Q 1 , Q 2 , . . . , Q n in a vacuum (Figure 1.42 ), the total electric field created by them at ¯ r can be obtained as 51 51 We assume that electric charges do not affect each other: i.e. the existence of an electric charge does not change the point properties of other electric charges or the electric fields created by them. This also means a point source creates the same electric field, whether there are other electric charges or not. ¯ D (¯ r ) = n i =1 Q i 4 π (¯ r − ¯ r i ) | ¯ r − ¯ r i | 3 , (1.182) where ¯ r i represents the location of the electric charge Q i . If there is a continuous volumetric distribution of electric charges as defined in Eq. ( 1.98 ), then the electric field created by this electric charge distribution (Figure 1.43 ) can be found via integration as 52 52 One can take the divergence of the electric flux density as ¯ ∇ · ¯ D (¯ r ) to derive the differential form of Gauss’ law for static cases, keeping in mind that the ¯ ∇ operator is applied to unprimed coordinates and using lim R → 0 ¯ ∇ · ˆ a R /R 2 = 4 πδ ( R ). ¯ D (¯ r ) = 1 4 π V q v (¯ r ) (¯ r − ¯ r ) | ¯ r − ¯ r | 3 d ¯ r .
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