Electrostatics PDF

Lines of electric field The great experimental physicist Faraday, in his revolutionary papers on electromagnetism, did not write down a single equation! How did he do it? He visualized the electric (and magnetic) fields. Lines of electric field The great experimental physicist Faraday, in his revo...

Lines of electric field The great experimental physicist Faraday, in his revolutionary papers on electromagnetism, did not write down a single equation! How did he do it? He visualized the electric (and magnetic) fields. Lines of electric field The great experimental physicist Faraday, in his revolutionary papers on electromagnetism, did not write down a single equation! How did he do it? He visualized the electric (and magnetic) fields. Lines of electric field The great experimental physicist Faraday, in his revolutionary papers on electromagnetism, did not write down a single equation! How did he do it? He visualized the electric (and magnetic) fields. Lines of electric field The great experimental physicist Faraday, in his revolutionary papers on electromagnetism, did not write down a single equation! How did he do it? He visualized the electric (and magnetic) fields. Lines of electric field Some basic rules and observations about field lines:  They never start or stop in empty space – they stop or start on a charge or extend to infinity.  They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there.  The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction. Lines of electric field Some basic rules and observations about field lines:  They never start or stop in empty space – they stop or start on a charge or extend to infinity.  They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there.  The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction. Lines of electric field Some basic rules and observations about field lines:  They never start or stop in empty space – they stop or start on a charge or extend to infinity.  They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there.  The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction. E density  Principle of Superposition F2 Electric Force on the charge, Q, due to the charges, q1, q2, … qi F1 Fi F = F1 +F2 + … Fi  Principle of Superposition Thus, if we can find the force on Q due to a single source charge q, we are, in principle, done !!  Then the rest is just a question of repeating the But is this so easy? same operation over and over, and adding it all up  The force on Q depends on the separation distance r between the charges (Fig. 2),  It also depends on both their velocities and on the acceleration of q.  Moreover, it is not the position, velocity, and acceleration of q right now that matter: What concerns Q is the position, velocity, and acceleration q had at some earlier time.  At this stage, we shall consider the special case of electrostatics in which all the source charges are stationary (though the test charge may be moving). Coulomb’s law and the principle of superposition constitute the physical input for electrostatics—the rest, except for some special properties of matter, is mathematical elaboration of these fundamental rules. O  E is called the electric field of the source charges. It is a function of position (r), because the separation vectors ri depend on the location of the field point P.  It makes no reference to the test charge Q.  The electric field is a vector quantity that varies from point to point and is determined by the configuration of source charges;  Physically, E(r) is the force per unit charge that would be exerted on a test charge, if you were to place one at P Application of Principle of Superposition to Calculate Electric Field Problem Find the electric field at a distance z above the center of a circular loop of radius r that carries a uniform line charge λ. The figure illustrates the situation. Answer First draw the electric field lines as shown in the figure below to get an idea about the situation.  The horizontal components will cancel each other out dE1 cos dE2 cos  The vertical components, however, will be added dE1 dE2 dE1 sin dE2 sin dl dl  Electric Flux  The strength of an electric field is proportional to the number of field lines per area.  The number of electric field lines that penetrates a given surface is called an “electric flux,” which we denote as ΦE.  The electric field can therefore be thought of as the number of lines per unit area.  Electric Flux  The strength of an electric field is proportional to the number of field lines per area.  The number of electric field lines that penetrates a given surface is called an “electric flux,” which we denote as ΦE.  The electric field can therefore be thought of as the number of lines per unit area. Electric Flux dA E Definition of an element of electric flux: E·dA Electric field E Oriented area element dA The direction of dA is always perpendicular to the surface. In this example, E·dA is (E)(dA) cos 145°. Electric Flux Flux E E E dA dA dA E Definition of an element of electric flux: E·dA Electric field E Oriented area element dA The direction of dA is always perpendicular to the surface. In this example, 0 ≤ E·dA ≤ (E)(dA). Gauss’s law q   E  E  dA  0 The total flux ФE (the sum of all the fluxes) through a closed surface equals the total charge inside the surface divided by ε0. Gauss’s law q   E  E  dA  0 The total flux ФE (the sum of all the fluxes) through a closed surface equals the total charge inside the surface divided by ε0. If there is no charge inside a closed surface, then all the field lines entering must also leave! Then the total flux is zero. Gauss’s law q   E  E  dA  0 ФE > 0 Gauss’s law q   E  E  dA  0 ФE = 0 Gauss’s law q   E  E  dA  0 ФE < 0  Essence of Gauss’s law  The field strength  is proportional to the density of field lines (the number per unit area),  Therefore, E · da  is proportional to the number of lines passing through the infinitesimal area da.  The dot product picks out the component of da along the direction of E, it is the area in the plane perpendicular to E that we have in mind when we say that the density of field lines is the number per unit area.  This suggests that the flux through any closed surface is a measure of the total charge inside. For the field lines that originate on a positive charge must either pass out through the surface or else terminate on a negative charge inside (Fig. 16a).  On the other hand, a charge outside the surface will contribute nothing to the total flux, since its field lines pass in one side and out the other (Fig. 16b). where Qenc is the total charge enclosed within the surface. This is the quantitative statement of Gauss’s law. This equation carries the same message as the previous one, it is Gauss’s law in differential form Symmetry is crucial to this application of Gauss’s law. As far as I know, there are only three kinds of symmetry that work Application of Gauss’s Law to Calculate Electric Field Example: Electric field due to a point charge q Let’s use Gauss’s law to derive the electric field of a point charge q. Construct a gaussian sphere of radius r centered on the point charge. By symmetry, E must be radial, and E must be constant and so   E  E  dA  ( E )(4 r 2 ). r Now we apply Gauss’s law, q   E  4E r 2 , 0 and immediately obtain E = q/4πε0r2, which is Coulomb’s law. r R Continued …. r R Example 5. An infinite plane carries a uniform surface charge σ. Find its electric field. Gaussian Solution: Draw a “Gaussian pillbox”, extending equal distances above pillbox and below the plane (Fig. 22). Apply Gauss’s law to this surface: Applications Gauss’s law makes it easy to obtain the electric field when there is symmetry to help us. For example, what is the electric field of an infinite rod with uniform charge density λ? Let the rod be the axis of a cylinder of radius ρ and height h; by symmetry, we have h q  0 0    E  E  dA  ( E )(2h) , E  so E . E 2 0 Applications Similarly, we can use Gauss’s law to calculate the electric field of a plane with uniform charge density σ. If we take a box of any cross-sectional area A and enclose part of the plane in the box, the electric flux will be zero on the sides of the box and 2EA through the top and bottom of the box. Thus E A q   0 0    E  E  dA  2 EA , so E  2 0.