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.

How do inkjet printers work?

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:

Both act in a vacuum.

Both are central and conservative.

Both obey an inverse-square law (both are inversely proportional to square of r).

Both propagate with finite speed c.

Differences between electrostatic and gravitational forces:

Electrostatic forces are much greater than gravitational forces (by about 1036 times).

Gravitational forces are attractive for like charges, whereas electrostatic forces are repulsive for like charges.

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.

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.

Positive and negative charges.

Try this appletOr this one A configuration with two charges equal in magnitude but of opposite sign is called an electric dipole.

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.
The Electric Field Due to an Electric Dipole

(a) An electric dipole. The electric field vectors E(+) and E(-) at point P on the dipole axis result from the dipole's two charges. Point P is at distances r(+) and r(-) from the individual charges that make up the dipole. (b) The dipole moment p of the dipole points from the negative charge to the positive charge.

The 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).

In situations when z » d, the term on the right will reduce to give us:

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.

The Electric Field Due to a Line of Charge

A ring of uniform positive charge. A differential element of charge occupies a length ds. This element sets up an electric field dE at point P. The component of dE along the central axis of the ring is dEcos(theta).

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: 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:

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):

To add all of the differential components along the ring, we simply integrate both sides of this equation from s = 0 to s = 2πR:

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:

Sample Problem

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?

(a) A plastic rod of charge -Q is a circular section of radius r and central angle 120 degrees; point P is the center of curvature of the rod. (b) A differential element in the top half of the rod, at an angle theta to the x axis and of arc length ds, sets up a differential electric field dE at P. An element ds', symmetric to ds about the x axis, sets up a field dE' at P with the same magnitude. (c) Arc length ds makes an angle dtheta about point P.

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.

An element: 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

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

The direction of dE is toward ds because charge dq is negative. Symmetric partner: 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. Summing: thus, 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

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

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:

(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.) Charge density: to 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 2πr/3, and its linear charge density must be

Substituting this into the previous equation and simplifying give us

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

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. 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)/r2, where r = d - x is the distance from the charge at x to the point in which we are interested.

Evaluating the integral leads to

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. The Electric Field Due to a Charged Disk

A disk of radius R and uniform positive charge. The ring shown has radius r and radial width dr. It sets up a differential electric field dE at point P on its central axis.

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 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

We can now find E by integrating the above equation over the surface of the disk from r = 0 to r = R:

If we let Rà∞ while keeping z finite, or if we let zà0 while keeping R finite, this equation reduces to

A Point Charge in an Electric Field

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.

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

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 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. Inkjet Printing

The essential features of an inkjet printer. Drops are shot out from generator G and receive a charge in charging unit C. An input signal from a computer controls the charge given to each drop and thus the effect of field E on the drop and the position on the paper at which the drop lands. About 100 tiny drops are needed to form a single character.

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.

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.

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.

A Dipole in an Electric Field

(a) An electric dipole in a uniform external electric field E. Two centers of equal but oposite charge are separated by distance d. The line between them represents the rigid connection. (b) Field E causes a torque tau on the dipole. The direction of the torque is into the page, as represented by the arrow tail symbol.

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 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.

Potential Energy of an Electric Dipole 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

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

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

Microwave Cooking

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:

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.

electric fieldsand their properties.The Electric FieldThe

electric fieldis 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 apositive point test charge q0at that point and calculating the ratio of the vector forceFacting on the test charge to the magnitude of this charge:The direction of

Eis the same as that ofF, and the test chargeq0is 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

Eand the gravitational fieldg, so sometimes mass is called "gravitational charge".Similarities between electrostatic and gravitational forces:

- Both act in a vacuum.
- Both are central and conservative.
- Both obey an inverse-square law (both are inversely proportional to square of r).
- Both propagate with finite speed c.

Differences between electrostatic and gravitational forces:Electric Field LinesMichael 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, calledelectric 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.

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 ChargeTo find the electric field due to a point charge

qat any point a distancerfrom the point charge, we put a positive test chargeq0at that point, and using Coulomb’s law obtain the electrostatic force acting onq0:The direction of

Fis directly away from the point charge ifqis positive and directly toward the point charge ifqis 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.

The Electric Field Due to an Electric Dipole

The figure shows two charged particles of magnitude

qbut of opposite sign, separated by a distanced. We now set off to find the electric field due to this dipole at pointP, a distancezfrom the midpoint of the dipole and on the axis through the particles (the dipole axis).In situations when

z»d, the term on the right will reduce to give us:The product

qdis the magnitude of the vector quantity known as theelectric dipole moment p, which always points from the negative to the positive end of the dipole.The Electric Field Due to a Line of ChargeWe 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 radiusRwith a uniform positive linear charge densityλaround its circumference. The pointP, a distancezfrom 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 lengthdswith charge of magnitudedq:This differential charge sets up a differential electric field

dEat pointP, which is a distancerfrom the element. Treating this element as a point charge we can express the magnitude ofdEas:dEis at angleθto the central axis (thezaxis) 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):To add all of the differential components along the ring, we simply integrate both sides of this equation from

s= 0 tos= 2πR:If the charge on the ring is negative, the magnitude of the electric field at

Pwould be given by the same equation, but its vector would point toward the ring instead of away from it. In situations whenz»R, the term on the right will reduce to give us:Sample ProblemThe figure shows a plastic rod having a uniformly distributed charge

–Q. The rod has been bent in a 120° circular arc of radiusr. We place coordinate axes such that the axis of symmetry of the rod lies along thexaxis and the origin is at the center of curvaturePof the rod. In terms ofQandr, what is the electric fieldEdue to the rod at pointP?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.

consider a differential element having arc lengthAn element:dsand located at an angleθabove thexaxis (b). If we letλrepresent the linear charge density of the rod, our elementdshas a differential charge of magnitudeour element produces a differential electric fieldThe element’s field:dEat pointP,which is a distancerfrom the element. Treating the element as a point charge, we can express the magnitude ofdEasThe direction of

dEis towarddsbecause chargedqis negative.our element has a symmetrically located (mirror image) elementSymmetric partner:ds’in the bottom half of the rod. The electric fielddE’set up atPbyds’also has the magnitude given by the above equation, but the field vector points towardds’as shown in the figure (b). If we resolve the electric field vectors ofdsandds’intoxandycomponents as shown, we see that theirycomponents cancel (because they have equal magnitudes and are in opposite directions). We also see that theirxcomponents have equal magnitudes and are in the same direction.thus, to find the electric field set up by the rod, we need sum (via integration) only theSumming:xcomponents of the differential electric fields set up by all the differential elements of the rod. We can write the componentdExset up bydsasThis equation has two variables,

θands.Before we can integrate it, we must eliminate one variable. We do so by replacingds,using the relationin which

dθis the angle atPthat includes arc lengthds(c). With this replacement, we can integrate the above equation over the angle made by the rod atP,from θ=-60° to θ=60°; that will give us the magnitude of the electric field atPdue to the rod:(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.)to evaluateCharge density: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 2πr/3, and its linear charge density must beSubstituting this into the previous equation and simplifying give us

The direction of

Eis toward the rod, along the axis of symmetry of the charge distribution. We can writeEin unit-vector notation asThe Electric Field Due to a Continuous Charge Distribution

Image the curved rod from the above example problem straightened out into a rod of finite length

Land a point in space a distancedfrom the rod, as in the above picture. If we takex= 0 at the center of the rod, and let the positivexaxis be to the right then we really only need to do a single integral to calculate the total field along thexaxis. If we then break the rod down into short pieces of lengthdx, each with chargedq, we then havedq= (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)/r2, wherer=d-xis the distance from the charge atxto the point in which we are interested.d, the expression in brackets gets smaller and smaller. This is because the two fractions become nearly the same asdincreases, and tend to cancel out. This makes sense, since the field should die off for larger values ofd, with distance squared. It is also worth noting that for lagerdthis distribution looks like a point charge.The Electric Field Due to a Charged DiskNext 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 radiusRthat has a positive surface charge of uniform densityσon its upper surface. Of interest is the electric field at pointPa distancezfrom 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. Sinceσis the charge per unit area, the charge on the ring iswhere

dAis the differential area of the ring. Since we already know how to deal with rings, we can use previous equations to obtainWe can now find

Eby integrating the above equation over the surface of the disk fromr= 0 tor=R:If we let

Rà∞ while keepingzfinite, or if we letzà0 while keepingRfinite, this equation reduces toA Point Charge in an Electric FieldHaving discussed electric fields due to charged objects, we now come full

ringin 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.Given a particle with charge

qinside an electric field of magnitudeEproduced by other charges, the electrostatic force on that particle is given byIt is worth mention that

Erepresents the external field (from the perspective of our particle in question); it does not factor in the particle. The above equation indicates thatThe electrostatic force

Facting on a charged particle located in the external electric fieldEhas the direction ofEif the chargeqof the particle is positive and has the opposite direction ifqis negative.Inkjet PrintingThe 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.

The figure shows a negatively charged drop moving between two conducting deflecting plates, between which a uniform, downward-directed electric field

Ehas 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 ofEand the chargeqof the drop.In practice,

Eis held constant and the position of the drop is determined by the chargeqdelivered 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.A Dipole in an Electric FieldThe 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 distanced. The dipole momentpmakes an angleθwith fieldE. 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 distancexfrom the center of mass, the other one must be a distance (d-x) from the center of mass. Thus we can calculate torque asThe torque acting on the dipole tends to rotate it into the direction of field

E, thereby reducingθ. In this case the direction is clockwise as indicated by (b), which gives torque negative sign.Potential Energy of an Electric DipoleThe dipole has its least potential energy

Uwhen it is in the stable equilibrium, that is whenpis in the same direction asE, which makesθ= 0º. The potential energy is greatest whenpis in the opposite direction ofE, makingθ= 180º. Choosing the zero point for potential energy atθ= 90º leads us toWhen a dipole rotates from an initial angle

θito another angleθf, the workWdone on the dipole by the electric field isIf the change in orientation is caused by an applied torque (due to an external agent), then the work

Wadone on the dipole by the applied torque is the negative of the work done on the dipole by the field:Sample Problem

Microwave CookingWater 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:

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.