Chapter 2 THE ELECTRC FELD ntroduction Discrete charges: Electric dipole Continuous charge distributions Flux of a vector field Flux of an electric field for a spherical Gaussian surface concentric to a point charge Solid angle Flux of an electric field for an arbitrary Gaussian surface enclosing a point charge Gauss law Summary NTRODUCTON Last lecture introduced the concept of electric charge and the force between two charges which is described by Coulomb s law. Coulomb s Law: F 10 1 1 0 4 0 10 2 cr 10 (2.1) t was stated that electric forces add vectorially, that is, the Principle of Superposition is obeyed. Principle of Superposition F 0 X F 0 (2.2) Because the electrostatic force depends only on position and time, it obeys superposition, and is proportional to the probe charge, it is possible to factor out the probe charge from the force which allows defining the concept of the electric field E Electric Field E F0 0 (2.3) Figure 1 E field at point P for n discrete point charges. The electric field is a vector field; at each point in space it has both a magnitude and direction describing the direction and magnitude of the electric force exerted on a unit positive charge. The electric field lines radiate from a positive charge and end at a negative charge. Superposition of electric forces implies superposition of electric fields. Using Coulomb s law plus superposition gives that the total electric field produced by charges for the point at position 0, is; E 0 X E 0 X 1 1 4 0 0 2 cr 0 (2.4) where the symbol P 1 means summation over 1 to, and the unit vector cr 0 points from 0. Once the electric fieldisknown,thenatanyposition it is possible to compute the electrostatic force on any charge q since F E (2.5) Knowing the electrostatic force and Newton s laws of motion allows computing the resultant motion of charged objects. As will be discussed later, electromagnetic radiation, such as light, is a manifestation of oscillating electric and magnetic fields. The ability to calculation electric fields is occasionally required in the biomedical field. For example, electric fields feature prominently in molecular binding and some animals, such as sharks, are able to locate food via detecting electric fields. Therefore it is useful in P114 to discuss the best means for calculation of electric fields for both discrete and continuous charge distributions. There are three approaches for calculating the electric field that use; (1) Coulomb s law, 2) Gauss law, or (3) electrostatic potential. These three methods each have advantages for calculation of electric fields for different systems. This discussion also will introduce you to the important concept of electric flux and Gauss Law which supersedes Coulomb s Law. 7
Figure 2 The electric dipole comprises two equal and opposite charges separated by a distance d. The electric dipole moment p points from -q to +q. Figure 3 dipole. The electric field distribution for an electric DSCRETE CHARGE DSTRBUTONS; ELECTRC DPOLE The electric field due to a system of point charges can be calculated using Coulomb s law by calculating the field due to each charge independently and then taking the vector sum using the Principle of Superposition as described above. The most important and simplest example of a discrete charge distribution is the electric dipole, which comprises two charges with equal magnitude and opposite sign, ±, separated by a distance d. The dipole is defined to have a electricdipolemoment p d pointing from to +. The net electric field is effectively the difference between the electric field due to two equal-magnitude and opposite sign point charges that are displaced in location. The electric dipole field was demonstrated using grass seeds and is illustrated in figure 3. Coulomb s law can be used to compute the electric field for an electric dipole. However, this field is complicated except for points on the parallel, or perpendicular axes to the dipole moment p Therefore, let us compute the electric field for these two cases where the mathematics is easy. Consider a coordinate system whose origin is at the center of the dipole with the axis along the direction of p and the axis perpendicular to p, in order to compute the electric field along these two axes Electric field at points perpendicular to dipole Since points on the perpendicular axis are equidistant from the two charges, then the components of the E field from each charge are equal and opposite, thus they cancel. Therefore we are left with a net component in the - b i direction that is twice the component of the field from each charge. E b 2 i 4 0 2 cos +b j0 (2.6) where 2 2 +( 2 )2. Since cos 2 and p b i then E b p i 4 0 3 4 0 3 (2.7) Electric dipole field at points on the dipole axis: The electric field from each of the charges points along the b i axis, but in opposite directions. Thus the net field is: E k b 1 i ( 4 0 ( 1 2 )2 ( + ) 2 )2 b 2 i 4 0 ( 2 ( 2 )2 ) 2 2 p 4 0 ( 2 ( 2 )2 ) 2 (2.8) Note that for distances, then the above formula simplifies to: E k 2 p 4 0 3 (2.9) Note that the E fieldfortheelectricdipoleisnotradial. The 1 radial dependence occurs because, to first 3 2 order, the 1 Coulomb fields for each of the two point charges partially cancel. Theelectricdipoleisofconsiderableimportancein physics. When an atom is in an external electric field, 8
Figure 4 Water molecule, H 2 O, is a permanent electric dipole. Both electrons are shared with the oxygen atom creating a strong electric bond. the positive nucleus and negative electron cloud are displaced leaving the atom with a net electric dipole moment. Some molecules, like water, have permanent electric dipole moments because one of the atoms binds the electrons more than do the companion atoms. The electric fields produced by aligned molecular electric dipoles, such as the water molecule, can drastically change the electric field inside non-conducting materials, which are called dielectrics. This has important consequences in biological matter as will be discussed later. Figure 5 Representation of continuous charge distributions confined to a line, surface, and volume. Linear charge density is the charge per unit length CONTNUOUS CHARGE DSTRBUTONS n most practical applications one is dealing with 10 20 atomic charges and it is impractical and unnecessary to try to take the vector sum of the fields from each of these individual charges. For distances large compared with the size of the atom, one can use the concept of a continuous charge distribution. Depending on the problem one defines volume, surface and linear charge densities in solving for electric fields, where: Volume charged density, is the charge per unit volume 1 (2.10) Surface (areal) charge density, isthechargeper unit area S (2.11) 1 To avoid confusion with the use of the letter for electric potential, the Greek character will be used to designate volume (2.12) For continuous charge distributions, the summation over discrete charges transforms into an integral over the continuous charge distribution. Thus the field at the point P due to charges at P is: E 1 4 0 2 dr 0 (2.13) 0 0 where br is the unit vector pointing from the charge element to the field point. The element of charge can be written in terms of the charge density and volume element, that is,. Thus the integral can be written as: E 1 4 0 1 E 4 0 E 1 4 0 br Volume charge (2.14) 2 br Surface charge (2.15) 2 br Linear charge (2.16) 2 The above integrals can be solved analytically for some uniform charge distributions, but a computer is needed to solve the general problem. 9
Figure 6 The electric field due to a uniform linear charge density, of length 2L at a distance R. Figure 7 Corona discharge around a high-voltage power line is due to the high electric field which ionizes the air resulting in electric breakdown. Example 1: E on the bisector of a finite uniform linear charge: Consider the field in the x-y plane at a perpendicular distance from the line charge of linear charge density C/m and subtending angle between ± 0 as shown in figure 6. The electric field then is given by the integral 1 E 4 0 br (2.17) 2 The unit vector br b i cos b j sin, while tan and cos Therefore and (tan ) cos 2 (2.18) 1 2 cos2 2 (2.19) Substitution of these leads to; 0 E ( 4 0 b i cos b j sin ) (2.20) 0 The limits ± 0 are the maximum values corresponding tothetwoendsofthelinecharge,wheretan 0. The integral simplifies to E 4 0 (b i sin + b j cos ) 0 0 (2.21) E 2 0 b i sin 0 (2.22) t is obvious from symmetry that the y components cancel as is also given mathematically. Note that for cases where À then 0 90 0 and sin 0 1 Figure 7 shows the corona discharge around a highvoltage power line which illustrates the electric field for alinecharge. Figure 8 Electric field along axis of a uniform circular ring of charge Example 2: E field on the axis of a circular ring charge: As shown in figure 8, the axial component of the electric field due to the charge element at a point along the symmetry axis is given by 1 4 0 2 cos 1 4 0 2 4 0 3 (2.23) Thus the total electric field along the axis is given by the integral 4 0 3 1 4 0 3 (2.24) By symmetry it is obvious that on the axis of the ring charge the perpendicular component of the electric field is zero. 10
Figure 9 Electric field on the axis of a uniform circular disk of charge Example 3; E field on the axis on a uniform circular disk of charge: Consider a uniform circular disk that has an areal charge density C/m 2 Take an infinitessimal concentric ring of charge radius and width then the total ring charge is 2 (2.25) From the previous example we know that the axial electric field due to an infinitessimal ring charge of radius to be 1 2 (2.26) 4 0 3 But 2 2 + 2 therefore the net electric field is given by 1 2 4 0 0 ( 2 + 2 ) 3 2 2 0 0 ( 2 + 2 ) 3 2 Ã! 1 (2.27) 2 0 ( 2 + 2 ) 1 2 Note that when then 2 0 which is a constant electric field pointing perpendicular away for the disk for a positive charge distribution. n general, the direct integration of Coulomb s Law to calculate the electric field is non-trivial, usually it requires a computer to evaluate. The above method uses brute force integration to compute the electric field. An elegant and much simpler method, based on Gauss Law can be exploited to determine the electric field distributions for systems possessing spatial symmetry. These are based on the concept of the scalar quantity called the flux of a vector field. Thus at this stage it is necessary to introduce the concept of flux and Gauss law. Figure 10 surface. Flux of vector field through an element of FLUX OF A VECTOR FELD Vector fields, suchasanelectricfield, magnetic field, fluid flow, etc., have both a magnitude and direction that depend on the position. Vector fields can be characterized using two important concepts that will be introduced in this and following lectures, namely, flux and circulation. For example, the rate of loss of water out of a draining bath tub can be related to the liters/sec flowing down the drain plus the circulation or angular momentum carried away by the flowing water. The concepts of flux and circulation make it possible to express Coulomb s Law in a general form that is more powerful. This lecture will lead to Gauss s Law which relates the net flux of the electric field over a closed surface to the net charge enclosed by the closed surface. First it is necessary to introduce the concept of flux Φ Flux, Φ is the normal component of the vector field, F through any surface. t is given by the integral over the surface. Φ F S (2.28) where the scalar product, also called dot product, was defined in P113, namely F S cos where is the angle between the vector F and the normal to the surface element S Note that the flux is the normal component to the surface of the vector field, it excludes the component along the surface; the surface component figures in the definition of circulation. The most interesting case is when the surface is closed, this is called a Gaussian surface. For a closed surface the flux of the vector field F is written as: Φ F S (2.29) where the symbol means a closed surface. By definition,foraclosedsurfacethesurfacevector S is 11
Figure 11 A gun firing 10 bullets/sec surrounded by an inner closed surface A and outer closed surface B. assumed to point outward. To illustrate the usefulness of flux, consider that your friendly NRA member uses his assault weapon as a source of bullets that are sprayed out at a rate of 10 bullets per second. The spray of bullets is a vector field, F which at any location, has a magnitude for the rate of bullets per unit area in the direction of the velocity vector. The flux Φ of bullets is the integral of the rate of bullets passing through some defined surface. Note that the flux is a scalar number equal to the number of bullets fired per second. For a closed surface that encloses the gun, and assuming that the bullets are not stopped, then the total flux through this closed surface must equal the source strength, that is, the number of bullets fired per second by the gun independent of the direction that the bullets spray or theshapeoftheclosedsurface. Figure 11 shows two closed surfaces that enclose the machine gun, firing 10 bullets per second, an inner surface A and an outer surface B that encloses both the gun and surface A. f the bullets do not hit anything, then 10 bullets per second will pass through both surfaces A and B independent of the shape of these closed surfaces. Of course there will be a delay between when first pull the trigger and when the bullets reach the surfaces. The brute force method to solve this system is to calculate the trajectory of each bullet using Newton s Law and requires input of the velocty vector, mass, size, angular momentum, and air drag for each bullet and then use superpostion. The usefulness of flux is that one can solve many problems without knowledge of the detailed motion of each bullet. For example, if the number of bullets is conserved, then the net flux out of any closed surface must equal the source strength contained by that closed surface, that istherateatwhichthegunfires bullets. f the enclosed volume does not contain a source of bullets, then the net flux must equal zero. For example, consider the volume lying between surfaces A and B. The net flux entering this volume through surface A is equal to the net flux exiting through surface B, that is the net flux 12 Figure 12 A concentric spherical surface of radius R enclosing a point charge q. out of this volume is zero. The volume enclosed by the closed surfaces A and B does not contain a source of bullets which is consistent with the zero net flux. f the intended victim lies within the volume between surfaces A and B, then the net flux flowing outwards will be reduced by the rate that bullets hit the victim, that is, the victim is a sink for bullets. Of course the victim could also become a source by shooting back with another gun. What is useful about the concept of flux is that it relates the basic property of source strength to net flux without getting bogged down in the gory details such as marksmanship, corpses, spilled blood, etc. The concept of flux is especially useful for dealing with vector fields like fluid flow, electric, and magnetic fields. FLUX OF ELECTRC FELD FOR A SPHERCAL GAUSSAN SURFACE CONCENTRC TO PONT CHARGE Considerthesimplecaseofapointcharge at the center of a concentric spherical Gaussian (closed) surface. For this case the field is uniform and normal to the spherical surface, that is, E 1 br (2.30) 4 0 2 Writing the spherical surface area element as ds br then the flux of out of the spherical surface, of radius, is: Φ Φ 4 0 2 E S (2.31) (br br) (2.32)
Since br br 10 and since the area of a sphere is 4 2, then Φ 4 0 2 42 (2.33) 0 Thus the net flux out of the closed spherical Gaussian surface is Φ E S 0 (2.34) This is a simple demonstration of Gauss Law which states that the net outward flux of the electric field equals the enclosed charge times a constant. Note that this simple-minded derivation of Gauss law assumed; a) point charge b) concentric spherical Gaussian surface c) Coulomb s law For this special case the net electric flux is independent of the radius of the sphere. This occurs because the 4 2 1 of the surface area exactly cancels the 4 2 dependence in Coulomb s Law. The second feature of Coulomb s law, used in this derivation, is that the field points radially outwards and thus is normal to the concentric spherical surface. This result is consistent with Faraday s description of the electric field in terms of electric field lines; that is, the number of field lines intersecting a concentric sphere is a constant independant of the radius of the sphere. Next we need to show that Gauss Law is equally applicable independent of the shape of the Gaussian surface surrounding a point charge. To show this requires introduction of the concept of solid angle. SOLD ANGLE The concept of solid angle should be familiar to you. t is a measure of the apparent size of a surface area as seen when viewing from some location. t is the two-dimensional analog of the angle subtended by a line when viewed from a given location. t is easiest to understand solid angle by comparison with the concept of angle. n two dimensions, the infinitesimal angle subtended by a short line element at a distance r is given by the projection of the line element on a concentric circle divided by the radius r of the circle. That sin is,. The unit of angle is the radian. The analogous relation for three dimensions gives the definition of solid angle. Consider a surface area element S at a radius. Then the solid angle Ω is defined as: Ω S br 2 cos 2 (2.35) where is the angle between the normal to the surface element and unit vector br. The solid angle is the apparent area of the surface element projected onto a Figure 13 Solid angle subtended by an area element ds and analog to the angle subtended by a line element dl. concentric sphere, divided by the square of the radius of the sphere. The unit of solid angle is the steradian which is dimensionless. Thus the net solid angle subtended by some surface is given by Ω br S 2 (2.36) The sun and the moon happen to subtend almost identical solid angles at the earth even though there is a factor of 200 difference in size which is cancelled by a similar ratio of the distances from the earth. The solid angle of a closed surface is an important special case that will be used frequently. Point enclosed by a closed surface The integral for a closed surface surrounding a point is Ω br S 2 4 (2.37) since the surface area of a sphere is 4 2 then a complete sphere must subtend a total solid angle of 4 steradians. The vector S is always taken to point outwards for a closed surface. Note that any shape of a closed surface subtends 4 steradians relative to a point inside the closed surface, that is, at all angles the closed surface completely encloses the point independent of whether the enclosing surface is a sphere, a cube, or some arbitrary shape. Point external to closed surface For points lying outside the closed surface, the net solid angle is zero. This can be seen by dividing the closed surface into two halves having the same perimeter as seen from the point illustrated in the figure. Both halves subtend the same magnitude solid angle. However, since the vector S is always taken to point outwards, then the cosine has the opposite sign for the two halves and 13
Figure 14 Solid angle for a closed surface, a) enclosing the point, and b) not enclosing the point. thus the integral of the solid angle over the external closed surface is zero since the contributions from the two halves cancel. FLUX OF ELECTRC FELD FOR AN ARBTRARY GAUSSAN SURFACE ENCLOSNG A PONT CHARGE Since the electric field for a point charge is given by 1 E br (2.38) 4 0 2 For an arbitrary closed surface enclosing the point charge the net flux is Φ E S Ã br! S 4 0 2 (2.39) But the above discussion of the solid angle gave that br S Ω 2 (2.40) This is just the term in brackets, that is Φ Ω (2.41) 4 0 Also from above we have that for a closed surface enclosing the point charge the total solid angle, in steradians, is given by br S Ω 2 4 steradians (2.42) 14 Figure 15 A single point charge surrounded by a concentric closed spheruical surface and an irregular closed surface. This is true for any shaped closed surface surrounding the point. Thus this gives that the net flux of the electric field due to an enclosed point charge is given by Φ E S Ω (2.43) 4 0 0 where the 4 factors cancel. The above argument shows that the net flux out of any shaped Gaussian surface enclosing a point charge is independent of the shape of the Gaussian surface or the exact location of the point charge within the enclosed volume. The assumptions made in obtaining this result are: a) Point charge b) Coulomb s law. This is a much more powerful statement of Gauss Law in that the net flux is independant of the shape of the closed surface or the location of the enclosed point charge as long as it lies within the closed Gaussian surface. The above proof is consistent with Faraday s description of the electric field in terms of electric field lines; that is, the number of field lines intersecting a concentric sphere is constant independant of the radius of the sphere. Since the number of field lines is conserved then it can be seen from figure 15 that the net flux of electric field lines is the same for any closed arbitrary closed surface enclosing a given point charge which is what has just been proved mathematically. Note that the net electric flux is zero for any charge that lies external to the closed Gaussian surface as illustrated in figure 14.
GAUSS LAW The above proof that for a point charge the net flux out of a closed surface enclosing a point charge is independent of the shape of the Gaussian surface, now can be extended to give Gauss law by invoking the Principle of Superposition. Superposition can be used to integrate over the charge distribution throughout the volume enclosed by the closed Gaussian surface resulting in the final form of Gauss Law. The Principle of Superposition allows extension to arbitrary charge distributions. Consider charges within the enclosed surface. Each charge produces an field which add to produce the net electric field. That is: E E 1 + E 2 + E 3 + (2.44) Now since the flux Φ is just a number, that is, it is a scalar: Figure 16 Closed Gaussian surface enclosing the positivecharge(a),thenegativecharge(b)andboth charges (C). The net flux out of these surfaces is positive for A, negative for B, and zero for C. Φ E S E 1 S + E 2 S + Φ 1 + Φ 2 + Φ 3 + 1 + 2 + 3 + 0 0 0 (2.45) Thus one has that for an arbitrary distribution of charges, the net flux is: Φ Ã E X S 1 0! (2.46) Finally, one can write the sum over an arbitrary continuous charge density distribution by taking for the infinitessimal volume to get the final form of Gauss s Law. Gauss s Law: Φ E S 1 0 (2.47) This relates the net flux out of a closed Gaussian surface to the total charge lying within the enclosed volume. Note that the assumptions used to derive this are: Coulomb s Law Principle of Superposition Note that the two crucial aspects of Coulomb s law that lead to Gauss law are that the electric field for a point charge is: exactly proportional to 1 2, the field is radial. Gauss s law is a restatement of Coulomb s law in a less transparent but more useful form. Gauss Law actually is one of Maxwell s four laws of electromagnetism. Gauss Law is completely equivalent to Coulomb s law for electrostatics or for slowly moving charges. However, Gauss law is more general and applies to electric fields arising from rapidly moving and accelerating charges where Coulomb s law does not apply. Really one should derived Coulomb s Law from Gauss s Law. t is interesting to apply Gauss Law to the case of the electric dipole. As seen in figure 16, ifthegaussian surface encloses only the positive charge then the net flux out is positive, if the Gaussian surface encloses onlythenegativechargethenthenetflux out is negative, that is, the flux is flowing into the surface. f the Gaussian surface encloses both charges then the net flux out is zero. However, this does not imply that there is no electric field. Actually one has flux flowing inward in some locations and outwards at others, such that the net total is zero as seen in Figure 16. Gauss law gives you na general property of the electric field that always applies. However, it does not give you the detailed distribution of the electric field. However, if there is an obvious spatial symmetry then one can use intuition to infer closed surfaces for which the direction and magnitude of the electric field must be constant. This allows one to then use Gauss law to deduce the magnitude of that electric field in a trivially easy way. This will be discussed and applied in the next chapter. 15
SUMMARY Static electric fields calculated using Coulomb s Law: Discrete charges: E Continuous charges: X 1 1 4 0 0 2 cr 0 (2.4) 1 E 4 0 br (2.14) 2 The concept of the flux of a vector field for a closed Gaussian surface was introduced where Φ F S (2.28) Also the concept of solid angle was introduced where the solid angle subtended by some surface is br S Ω 2 (2.31) For a closed Gaussian surface this integral equals br S Ω 2 4 steradians (2.37) The concepts of flux and solid angle were combined to show that the net electric flux for an arbitrary closed Gaussian surface surrounding a point charge is given by Φ Ω 4 0 0 which is independant of the shape of the closed surface. Superposition now can be used to compute the flux for many charges enclosed in the closed surface. Ã Φ E! X S (2.46) 1 0 For a continuous charge distribution this can be written as Φ E S 1 (2.47) 0 This important rule immediately leads to the powerful and crucially important Gauss Law which will be derived and then used to derive electric fields for charge distributions during the next lecture. Reading assignment: Giancoli Chapter 21.6-22.2 16