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Current and Resistance
Magnetic Fields Due to Current
Induction and Inductance
Electromagnetic Oscillations and Alternating Current
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
and their properties.
How do inkjet printers work?
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
acting on the test charge to the magnitude of this charge:
The direction of
is the same as that of
, and the test charge
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
and the gravitational field
, 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 applet
Or this one
A configuration with two charges equal in magnitude but of opposite sign is called an
An electric dipole.
This quick animation demonstrates a three dimensional model of the pictures above (click
The Electric Field Due to a Point Charge
To find the electric field due to a point charge
at any point a distance
from the point charge, we put a positive test charge
at that point, and using Coulomb’s law obtain the electrostatic force acting on
The direction of
is directly away from the point charge if
is positive and directly toward the point charge if
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.
figure shows two charged particles of magnitude
but of opposite sign, separated by a distance
. We now set off to find the electric field due to this dipole at point
, a distance
from the midpoint of the dipole and on the axis through the particles (the dipole axis).
In situations when
, the term on the right will reduce to give us:
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
with a uniform positive linear charge density
around its circumference. The point
, a distance
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
with charge of magnitude
This differential charge sets up a differential electric field
, which is a distance
from the element. Treating this element as a point charge we can express the magnitude of
is at angle
to the central axis (the
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(
To add all of the differential components along the ring, we simply integrate both sides of this equation from
= 0 to
If the charge on the ring is negative, the magnitude of the electric field at
would be given by the same equation, but its vector would point toward the ring instead of away from it. In situations when
, the term on the right will reduce to give us:
The figure shows a plastic rod having a uniformly distributed charge
. The rod has been bent in a 120° circular arc of radius
. We place coordinate axes such that the axis of symmetry of the rod lies along the
axis and the origin is at the center of curvature
of the rod. In terms of
, what is the electric field
due to the rod at point
(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.
consider a differential element having arc length
and located at an angle
axis (b). If we let
represent the linear charge density of the rod, our element
has a differential charge of magnitude
The element’s field:
our element produces a differential electric field
which is a distance
from the element. Treating the element as a point charge, we can express the magnitude of
The direction of
our element has a symmetrically located (mirror image) element
in the bottom half of the rod. The electric field
set up at
also has the magnitude given by the above equation, but the field vector points toward
as shown in the figure (b). If we resolve the electric field vectors of
components as shown, we see that their
components cancel (because they have equal magnitudes and are in opposite directions). We also see that their
components have equal magnitudes and are in the same direction
hus, to find the electric field set up by the rod, we need sum (via integration) only the
components of the differential electric fields set up by all the differential elements of the rod. We can write the component
set up by
This equation has two variables,
Before we can integrate it, we must eliminate one variable. We do so by replacing
using the relation
is the angle at
that includes arc length
(c). With this replacement, we can integrate the above equation over the angle made by the rod at
from θ=-60° to θ=60°; that will give us the magnitude of the electric field at
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
we would then have discarded the minus sign.)
, 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
/3, and its linear charge density must be
Substituting this into the previous equation and simplifying give us
The direction of
is toward the rod, along the axis of symmetry of the charge distribution. We can write
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
and a point in space a distance
from the rod, as in the above picture.
If we take
= 0 at the center of the rod, and let the positive
axis be to the right then we really only need to do a single integral to calculate the total field along the
axis. If we then break the rod down into short pieces of length
, each with charge
, we then have
. The pieces are infinitesimally short and as such we can treat them as point charges. This allows us to use the expression (k
is the distance from the charge at
to the point in which we are interested.
Evaluating the integral leads to
Lastly for large values of
, the expression in brackets gets smaller and smaller. This is because the two fractions become nearly the same as
increases, and tend to cancel out. This makes sense, since the field should die off for larger values of
, with distance squared. It is also worth noting that for lager
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
that has a positive surface charge of uniform density
on its upper surface. Of interest is the electric field at point
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
is the charge per unit area, the charge on the ring is
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
by integrating the above equation over the surface of the disk from
= 0 to
If we let
∞ while keeping
finite, or if we let
0 while keeping
finite, this equation reduces to
A Point Charge in an Electric Field
Having discussed electric fields due to charged objects, we now come full
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
inside an electric field of magnitude
produced by other charges, the electrostatic force on that particle is given by
It is worth mention that
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
acting on a charged particle located in the external electric field
has the direction of
if the charge
of the particle is positive and has the opposite direction if
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
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
and the charge
of the drop
is held constant and the position of the drop is determined by the charge
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
. 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
. The dipole moment
makes an angle
. 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
from the center of mass, the other one must be a distance (
) 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
, thereby reducing
. 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
when it is in the stable equilibrium, that is when
is in the same direction as
, which makes
= 0º. The potential energy is greatest when
is in the opposite direction of
= 180º. Choosing the zero point for potential energy at
= 90º leads us to
When a dipole rotates from an initial angle
to another angle
, the work
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
done on the dipole by the applied torque is the negative of the work done on the dipole by the field:
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:
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?
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.
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