ELECTRIC+FIELDS

Inkjet printers have been around for a long time and we have all used them. But how do they work? What is the process that targets specific areas on a sheet of paper to help tiny droplets compose the final image? We shall find the answer in this chapter by exploring //electric fields// and their properties.




 * The Electric Field **

The //electric field// is a vector field which consists of a distribution of vectors, one for each point in the region around a charged object, such as a charged rod. The electric field can be defined at some point near the charged object by placing a //positive point test charge q0// at that point and calculating the ratio of the vector force //F// acting on the test charge to the magnitude of this charge:

The direction of //E// is the same as that of //F//, and the test charge //q0// is always positive. The SI unit for the electric field is the newton per coulomb (N/C). The test charge is used only to measure the electric field; the field itself is present regardless of the presence of the test charge (we assume that the presence of the test charge does not affect the charge distribution on the charged object and thus does not alter the electric field in question).

Coulomb's law, which describes the interaction of electric charges: is similar to the Newtonian gravitation law: This suggests similarities between the electric field //E//  and the gravitational field //g// , so sometimes mass is called "gravitational charge". Similarities between electrostatic and gravitational forces:

Differences between electrostatic and gravitational forces:
 * 1) Both act in a vacuum.
 * 2) Both are central and conservative.
 * 3) Both obey an inverse-square law (both are inversely proportional to square of r).
 * 4) Both propagate with finite speed c.
 * 1) Electrostatic forces are much greater than gravitational forces (by about 1036 times).
 * 2) Gravitational forces are attractive for like charges, whereas electrostatic forces are repulsive for like charges.
 * 3) There are no negative gravitational charges (no negative mass) while there are both positive and negative electric charges. This difference combined with previous implies that gravitational forces are always attractive, while electrostatic forces may be either attractive or repulsive.
 * 4) Electric charge is invariant while relativistic mass isn't.


 * Electric Field Lines **

Michael Faraday, who introduced the idea of electric fields in the 19th century, thought of the space around a charged object as filled with //lines of force//. These lines, called //electric field lines//, are imaginary, but they are useful in visualizing patterns in electric fields. · At any point, the direction of a straight electric field line or the direction of the tangent to a curved field line gives the direction of the electric field at that point. · The field lines are drawn so that the number of lines per unit area, measured in a plane that is perpendicular to the lines, is proportional to the magnitude of the electric field. Electric field lines extend from positive charge, where they originate, and toward negative charge, where they terminate.

[|Try this applet] [|Or this one] A configuration with two charges equal in magnitude but of opposite sign is called an //electric dipole//.



This quick animation demonstrates a three dimensional model of the pictures above (click [|here] )


 * The Electric Field Due to a Point Charge **

To find the electric field due to a point charge //q// at any point a distance //r// from the point charge, we put a positive test charge //q0// at that point, and using Coulomb’s law obtain the electrostatic force acting on //q0//:

The direction of //F// is directly away from the point charge if //q// is positive and directly toward the point charge if //q// is negative. The electric field vector is then:

To find the net electric field due to more than one point charge, we simply sum individual electric field vectors. <span style="font-family: 'Times New Roman','serif';"> <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> **<span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> The Electric Field Due to an Electric Dipole **

The <span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">figure shows two charged particles of magnitude //q// but of opposite sign, separated by a distance //d//. We now set off to find the electric field due to this dipole at point //P//, a distance //z// from the midpoint of the dipole and on the axis through the particles (the dipole axis).

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">In situations when //z// » //d//, the term on the right will reduce to give us:

<span style="font-family: 'Times New Roman','serif'; font-size: 105%;">The product //qd// is the magnitude of the vector quantity known as the //electric dipole moment p//, which always points from the negative to the positive end of the dipole.

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;"> We now consider the electric field due to a charge consisting of many point charges distributed over a line (or a curve). To do that, we are going to introduce a new quantity – //linear charge density λ//, whose SI unit is the coulomb per meter (C/m). The figure shows a thin ring of radius //R// with a uniform positive linear charge density //λ// around its circumference. The point //P//, a distance //z// from the plane of the ring along its central axis, is the point where we want to measure the electric field. Since the charge distribution is continuous, we cannot simply sum up the electric field vectors due to each point charge on the ring; instead we shall consider a small element on the ring of length //ds// with charge of magnitude //dq//: <span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">This differential charge sets up a differential electric field //dE// at point //P//, which is a distance //r// from the element. Treating this element as a point charge we can express the magnitude of //dE// as:
 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;">The Electric Field Due to a Line of Charge **

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">//dE// is at angle //θ// to the central axis (the //z// axis) and has components perpendicular and parallel to that axis. Due to symmetry, the perpendicular components will cancel out, and the parallel ones can be expressed using cos(//θ//):

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">To add all of the differential components along the ring, we simply integrate both sides of this equation from //s// = 0 to //s// = 2π//R//:

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">If the charge on the ring is negative, the magnitude of the electric field at //P// would be given by the same equation, but its vector would point toward the ring instead of away from it. In situations when //z// » //R//, the term on the right will reduce to give us:

<span style="font-family: 'Times New Roman','serif'; font-size: 12pt;">**Sample Problem** <span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">

The figure shows a plastic rod having a uniformly distributed charge //–Q//. The rod has been bent in a 120° circular arc of radius //r//. We place coordinate axes such that the axis of symmetry of the rod lies along the //x// axis and the origin is at the center of curvature //P// of the rod. In terms of //Q// and //r//, what is the electric field //E// due to the rod at point //P//?



<span style="font-family: 'Times New Roman',Times,serif; font-size: 105%;">Because the rod has a continuous charge distribution, we must find an expression for the electric fields due to differential elements of the rod and then sum those fields via calculus.


 * //<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">An element: //**<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;"> consider a differential element having arc length //ds// and located at an angle //θ// above the //x// axis (b). If we let //λ// represent the linear charge density of the rod, our element //ds// has a differential charge of magnitude

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">**//The element’s field://** our element produces a differential electric field //dE// at point //P,// which is a distance //r// from the element. Treating the element as a point charge, we can express the magnitude of //dE// as

<span style="font-family: 'Times New Roman','serif'; font-size: 105%;">The direction of //dE// is toward //ds// because charge //dq// is negative.
 * //<span style="font-family: 'Times New Roman','serif';">Symmetric partner: //** <span style="font-family: 'Times New Roman','serif'; font-size: 105%;"> our element has a symmetrically lo­cated (mirror image) element //ds’// in the bottom half of the rod. The electric field //dE’// set up at //P// by //ds’// also has the magnitude given by the above equation, but the field vector points toward //ds’// as shown in the figure (b). If we resolve the elec­tric field vectors of //ds// and //ds’// into //x// and //y// components as shown, we see that their //y// components can­cel (because they have equal magnitudes and are in oppo­site directions). We also see that their //x// components have equal magnitudes and are in the same direction <span style="font-family: 'Times New Roman','serif';">.
 * //<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">Summing: //**<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;"> t <span style="color: black; font-family: 'Times New Roman','serif'; font-size: 11pt;">hus, to find the electric field set up by the rod, we need sum (via integration) only the //x// compo­nents of the differential electric fields set up by all the dif­ferential elements of the rod. We can write the component //dEx// set up by //ds// as

<span style="color: black; font-family: 'Times New Roman','serif'; font-size: 11pt;">This equation has two variables, //θ// and //s.// Before we can integrate it, we must eliminate one variable. We do so by replacing //ds,// using the relation

<span style="color: black; font-family: 'Times New Roman','serif'; font-size: 11pt;">in which //dθ// is the angle at //P// that includes arc length //ds// (c). With this replacement, we can integrate the above equation over the angle made by the rod at //P,// from θ=-60° to θ=60°; that will give us the magnitude of the electric field at //P// due to the rod:

<span style="font-family: 'Times New Roman','serif'; font-size: 105%;">(If we had reversed the limits on the integration, we would have gotten the same result but with a minus sign. Since the integration gives only the magnitude of //E,// we would then have discarded the minus sign.)
 * //<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">Charge density: //** <span style="font-family: 'Times New Roman','serif'; font-size: 11pt;"> to <span style="color: black; font-family: 'Times New Roman','serif'; font-size: 11pt;">evaluate //λ//, we note that the rod subtends an angle of 120° and so is one-third of a full circle. Its arc length is then 2π//r///3, and its linear charge density must be

<span style="color: black; font-family: 'Times New Roman','serif'; font-size: 11pt;">Substituting this into the previous equation and simplifying give us

<span style="color: black; font-family: 'Times New Roman','serif'; font-size: 11pt;">The direction of //E// is toward the rod, along the axis of sym­metry of the charge distribution. We can write //E// in unit-vector notation as

The Electric Field Due to a Continuous Charge Distribution ** <span style="font-family: 'Times New Roman','serif'; font-size: 11pt;"> <span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">Image the curved rod from the above example problem straightened out into a rod of finite length //L// and a point in space a distance //d// from the rod, as in the above picture. <span style="font-family: 'Times New Roman',Times,serif; font-size: 110%;">If we take //x// = 0 at the center of the rod, and let the positive //x// axis be to the right then we really only need to do a single integral to calculate the total field along the //x// axis. If we then break the rod down into short pieces of length //dx//, each with charge //dq//, we then have //dq// = (//dx/////L//)//Q//. The pieces are infinitesimally short and as such we can treat them as point charges. This allows us to use the expression (k//*dq)/////r//2, where //r// = //d// - //x// is the distance from the charge at //x// to the point in which we are interested. <span style="font-family: 'Times New Roman',Times,serif;">Evaluating the integral leads to
 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;">
 * <span style="font-family: 'Times New Roman',Times,serif;">[[image:http://www.vias.org/physics/img/equ_bk4_125_1.gif]] ||

<span style="font-family: 'Times New Roman',Times,serif;">Lastly for large values of //d//, the expression in brackets gets smaller and smaller. This is because the two fractions become nearly the same as //d// increases, and tend to cancel out. This makes sense, since the field should die off for larger values of //d//, with distance squared. It is also worth noting that for lager //d// this distribution looks like a point charge. <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;">
 * <span style="font-family: 'Times New Roman',Times,serif;">[[image:http://www.vias.org/physics/img/equ_bk4_125_2.gif]] ||
 * The Electric Field Due to a Charged Disk**

Next we concentrate our attention on finding the electric field due to a charged disk. This prompts for a new quantity – //surface charge density σ//, whose SI unit is the coulomb per meter squared (C/m2). The figure shows a circular disk of radius //R// that has a positive surface charge of uniform density //σ// on its upper surface. Of interest is the electric field at point //P// a distance //z// from the disk along its central axis. This time we will simply divide the disk into differential concentric flat rings and add up their contributing factors through integration. Since //σ// is the charge per unit area, the charge on the ring is <span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">where //dA// is the differential area of the ring. Since we already know how to deal with rings, we can use previous equations to obtain

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">We can now find //E// by integrating the above equation over the surface of the disk from //r// = 0 to //r// = //R//:

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">If we let //R// à <span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">∞ while keeping //z// finite, or if we let //z// à <span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">0 while keeping //R// finite, this equation reduces to




 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;">A Point Charge in an Electric Field **

<span style="font-family: 'Times New Roman','serif';">Having discussed electric fields due to charged objects, we now come full //ring// in our discussion about charges inside electric fields. We will start by focusing on just one particle in an electric field, and then will discuss dipole behavior.

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">Given a particle with charge //q// inside an electric field of magnitude //E// produced by other charges, the electrostatic force on that particle is given by

<span style="font-family: 'Times New Roman','serif'; font-size: 105%;">It is worth mention that //E// represents the external field (from the perspective of our particle in question); it does not factor in the particle. The above equation indicates that <span style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: yellow; font-family: 'Times New Roman','serif'; font-size: 105%; mozbackgroundclip: -moz-initial; mozbackgroundinlinepolicy: -moz-initial; mozbackgroundorigin: -moz-initial;">The electrostatic force //F// acting on a charged particle located in the external electric field //E// has the direction of //E// if the charge //q// of the particle is positive and has the opposite direction if //q// is negative. <span style="font-family: 'Times New Roman','serif'; font-size: 105%;"> <span style="font-family: 'Times New Roman','serif';"> <span style="font-family: 'Times New Roman','serif';"> <span style="font-family: 'Times New Roman','serif'; font-size: 105%;">The need for high-quality, high-speed printing has caused a search for an alternative to impact printing, such as occurs in a standard typewriter. Building up letters by squirting tiny drops of ink at the paper is one such alternative.
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 110%;">Inkjet Printing **

The figure shows a negatively charged drop moving between two conduct­ing deflecting plates, between which a uniform, downward-directed electric field //E// has been set up. The drop is deflected upward according to the above equation and then strikes the paper at a position that is determined by the magnitudes of //E// and the charge //q// of the drop <span style="font-family: 'Times New Roman','serif';">.

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">In practice, //E// is held constant and the position of the drop is determined by the charge //q// delivered to the drop in the charging unit, through which the drop must pass before entering the deflecting system. The charging unit, in turn, is activated by electronic signals that encode the material to be printed.


 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;">A Dipole in an Electric Field **

The figure shows a dipole in a uniform external electric field //E//. For the purpose of this discussion, this dipole is a rigid structure with charges of equal magnitude but different sign at each end, separated by distance //d//. The dipole moment //p// makes an angle //θ// with field //E//. Electrostatic forces of equal magnitude act on each charge of the dipole in opposite directions, thus adding up to zero relatively to the center of mass. However, these forces produce net torque about the center of mass. Assuming that one of the charges is a distance //x// from the center of mass, the other one must be a distance (//d-x//) from the center of mass. Thus we can calculate torque as <span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">The torque acting on the dipole tends to rotate it into the direction of field //E//, thereby reducing //θ//. In this case the direction is clockwise as indicated by (b), which gives torque negative sign.

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">The dipole has its least potential energy //U// when it is in the stable equilibrium, that is when //p// is in the same direction as //E//, which makes //θ// = 0º. The potential energy is greatest when //p// is in the opposite direction of //E//, making //θ// = 180º. Choosing the zero point for potential energy at //θ// = 90º leads us to
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Potential Energy of an Electric Dipole **

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">When a dipole rotates from an initial angle //θi// to another angle //θf//, the work //W// done on the dipole by the electric field is

<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">If the change in orientation is caused by an applied torque (due to an external agent), then the work //Wa// done on the dipole by the applied torque is the negative of the work done on the dipole by the field:

** Sample Problem **

**<span style="font-family: 'Times New Roman','serif';">Microwave Cooking ** <span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">



<span style="font-family: 'Times New Roman','serif'; font-size: 11pt;">Water molecules are dipoles, abundant in most foods. When a microwave oven is turned on, the magnetron sets up a rapidly oscillating electric field within the oven which permeates the cooking chamber and any object contained within. Water molecules in the food or object try to align themselves with the rapidly changing electric field, causing friction which generates heat to warm up and cook the contents of the oven. Foods and objects with low moisture contents are not affected by microwaves as much as those with high content, though any object that contains water molecules will be affected by the electric field. If water molecules were not electric dipoles, microwaves would not cook or heat our food.

You can "cook" other thigs too:

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Example 1: A disk of radius 2.5 cm has a surface charge density of 5.3 µC/m2 on its upper face. What is the magnitude of the electric field produced by the disk at a point on its central axis at distance z = 12 cm from the disk?

Example 2: An electron is accelerated eastward at 1.80 X 109 m/s2 by an electric field. Determine the (a) magnitude and (b) direction of the electric field.