When studying electric field we must understand the concept of point charge.
We mean by point charge, a charge that can have positive or negative polarity
and, furthermore, we consider that the entire load is concentrated in a point of infinitesimal dimension, that is, of negligible dimension.
To study electric fields we will postulate a field model that describes how charges interact:
1. Some charges, which we will call source charges, alter the space around them by creating an electric fieldE^{→}.
2. Every isolated charge within an electric field experiences a forceF^{→} exercised by the field.
We must solve two tasks to make this a useful model for electrical interactions. First,
We must learn how to calculate the electric field for a source charge configuration.
Second, we must determine the forces exerted on a charge and its motion within the electric field.
Suppose the charge q experiences an electric forceF^{→} due to other charges. So at each point
of space acts a certain force on the charge q. This way we will define electric field E^{→} according eq. 71-01.
eq. 71-01
So we define electric fieldE^{→}
as a ratio between force and charge , so that the electric field unit
E^{→} is the newton por coulomb, or N/C.
The module of the electric field is called the electric field intensity.
We can think of using a proof charge q to check if there is any electric field
in a certain place in space. If the charge q experiences an electric force at that point in space, we say that there is an electric field E^{→}
acting at that point causing the forceF^{→}.
Observation: note that the proof charge q also produces an electric field, but
charges do not exert forces on themselves. Therefore, the charge q measures only the electric field generated by other charges.
From the above, it is concluded that the electric field is the causative agent of the force on the charge q.
So we can establish some basic ideas about field:
If q is positive , the electric field vector will point in the same direction of the electric force exerted on that charge.
The eq. 71-01 associates a vector with each point in space, that is, the electric field is a
vector field.
It should be clear that the electric field does not depend on the proof charge q . It is only used to check if there is an electric field in place and, if possible, to measure it. The existing electric field at the site is generated by the source charges.
After these settings we can rewrite eq. 71-01 so that the force is a function of the electric field, or:
eq. 71-02
This equation allows to find the force on a given charge
at a place in space where the electric field is known.
Electric Field Direction Convention
It is conventional that the direction of the electric field vector generated by charges follows the following criteria:
1. If the charge polarity is positive the electric field vector points radially
"out" of the load.
2. If the charge polarity is negative the electric field vector points radially to "inside" the charge.
In Figure 71-01 we can see the scheme corresponding to this convention.
To study Coulomb's law, let's first understand the constants that are used in the equation.
The electric charge unit in SI is Coulomb. On the other hand, the amount represented by the fundamental charge unit, usually designated by the letter "e", has a value equal to e = 1.6 x 10^{-19} C.
Another constant used in the calculations is the electrostatic constant K which in units SIK = 8.99 x 10^{9} N m^{2} / C^{2}. In this site we will adopt the value
rounded off this constant, or K = 9 x 10^{9} N m^{2} / C^{2}.
As already defined in chapter 3, regarding capacitors, we have the so-called constant of
vacuum permittivity , ε_{o}, known as epsilon zero and of a value equal to:
There is a relation between the constant K and the constant ε_{o} given by
equation below:
eq. 71-03
After these definitions we can write the equation that defines the interaction force between two particles
electrically charged. Like this:
eq. 71-04
It is worth mentioning some important observations about the Coulomb's law, listed below:
The force, in modulus, that the charge q_{1} exerts on q_{2} is exactly equal to the force that q_{2} exerts on q_{1}, with the same direction but the opposite way.
Coulomb's law applies only to point-like charges, which is an idealized object with charge and mass but no extension or size. If objects are much smaller than the separation between them, we may consider them
point-like charges.
When a charge is under the action of several forces, the resulting force is the vector sum of all forces.
Example 71-1
Source: Problem elaborated by the author of the site.
Be two charges, q_{1} = 10 µC e q_{2} = 20 µC separated by
a distance of 0.1 meter . Calculate the electrical force between these two charges.
Solution
As we know the values of the charges and their distance just apply the eq. 71-03, that is:
F = (10 x 10^{-6} x 20 x 10^{-6}) x 9 x 10^{9} / (0.1)^{2}
The electric field created by an electric charge is responsible for the existence of electric forces
acting on other charges. If we insulate a charge, it will not be acted upon by forces. Although it produces
an electric field, it does not take action from its own field.. The forces between charged particles are called action at a distance . Somehow the force is transmitted through space. Just like other forces, such as gravitational, magnetic, etc ..., are forces that obey the inverse square law of distance.
In the literature some models of fields produced by different situations with objects are studied.
electrically charged. We will study some of them.
From the equations 71-01 and 71-03 we can find an equation
that allows us to calculate the electric field generated by a punctual charge. If we have a
charge q_{1} in a certain place in space and we are interested in knowing the electric field produced by it, it is only natural that we use a second charge, here called q_{2}, to serve as a "probe" or "proof charge". Then the charge q_{1} will produce an electric field such that the charge
q_{2} will experience a force acting about it. This field is spelled out in
eq. 71-01 where F^{→} is the force on q_{2}.
Then using eq. 71-03, we arrive at the electric field equation generated by q_{1}, according to eq. 71-05.
eq. 71-05
In this equation we are representing the electric field as a vector quantity. We chose to represent the charge in a generic way by the letter q. And r^ represents the unit vector, that is, vector of length ONE with orientation from origin to point of interest. Other variables are already known to us.
This equation also works for negative electric charges, causing the direction of the electric field vector to be contrary to that of a positive charge, that is, the electric field vector points toward the charge.
As has been said, it is important to distinguish the charges that are the sources of an electric field from those that they experience and move under the influence of this electric field. Suppose the source of an electric field is a set of point charges q_{1}, q_{2}, q_{3}, ...
So the resulting electric field ^{→}E_{res}
at each point in space is the overlap of the electric fields generated individually by each set charge at that point. Therefore, the best way to express the resulting electric field vector is in vector form, or:
To study the electric field it is more appropriate to consider the charge as being continuous and
describe how it is distributed by the object. Thus, we can have a linear, superficial or volumetric distribution.
In the case of linear distribution, we can assume an object with the length being the predominant dimension and having a length L . So, if the object is electrified and we call the total charge of the object as Q , we can define the linear charge density, λ, as
eq. 71-07
Linear charge density is defined as the amount of charge per meter in length and its unit is the C/m
The superficial charge density, σ, is the amount of charge distributed over a
surface area A. And your unit is the C/m^{2}, that is, it is the amount of charge per
square meter and can be represented by
eq. 71-08
The volumetric charge density, ρ, is the amount of charge distributed in a
volume V. And your unit is the C/m^{3}, that is, it is the amount of charge per
cubic meter and can be represented by
We are interested in calculating the electric field produced by an electrically charged wire of length
L, at a point P, located on its axis of symmetry, as we can see in Figure 71-02.
We assume a charge dq located at a point along the wire and we want to calculate the field
dE at point P. Let's remember that dq = λ da and r^{2} = x^{2} + a^{2}.
So, using Coulomb's law, we have:
eq. 71-10
This is the value of dE . However, we want to calculate the electric field on the symmetry axis, that is, dE_{x}. In this case, by Figure 71-02, we can see that
eq. 71-11
Therefore, it is clear that we must calculate the component of dE in the x direction. Based still in Figure 71-02, we can write that
eq. 71-12
Thus, to obtain the eq. 71-11 we must join the eq. 71-10 with eq. 71-12.
Like the eq. 71-11 represents only the differential of the field that we want to calculate, so to get the value of the electric field E_{x}, we must integrate eq. 71-11 from - L to + L. Thus, we can write
eq. 71-13
Note that the integration variable is a. So we can reorganize the integral as follows:
eq. 71-14
Now we are going to focus on the integral solution only. Notice that we have, in the denominator,
1 + (a/x)^{2} and this suggests a transformation of variables of the type tangent, because by Figure 71-02 we see that tan θ = a / x . So, a = x tan θ.
Therefore, da = x sec^{2} θ dθ.Furthermore, we know that
1 + tan^{2} θ = sec^{2} θ and making the substitution in the denominator we find the value of sec^{2} θraised to power 3/2 which results in
sec^{3} θ. Making the substitution in the numerator and denominator we find the
eq. 71-15.
eq. 71-15
But note that the cosine function can be expressed as the inverse of the secant function. In addition, we can take x out of the integral. Then it is possible to write:
eq. 71-16
And since the integral of the cosine is the sine function, and by Figure 71-02 we can see that sen θ = a/r = a / (a^{2} + x^{2})^{1/2} . Thus, replacing this value in eq. 71-16 we get
eq. 71-17
And finally, making the substitution we find the final equation that determines the
electric field value on the wire symmetry axis.
eq. 71-18
With this result we can make some considerations to verify its validity. Note that
the numerator of eq. 71-18 represents the total charge Q = 2 & lambda; L of the wire,
because the wire length is 2 L. We will take the limit when x → ∞.
In this case, x >> L and, therefore, in the denominator we have x^{2}. Therefore, the equation is reduced to Coulomb's law
for a punctual charge which represents a very reasonable behavior of the equation. On the other hand, we can analyze how it behaves by making the limit when x → 0. In this case, the equation
boils down to:
eq. 71-19
And this equation is exactly the equation that determines the electric field of a wire with an
infinite length. Therefore, the eq. 71-18 can be interpreted for various situations.
For an electrically charged disc with a surface charge density, σ, uniform, let's calculate the electric field it produces at a point P located along the z axis. The situation that we are going to analyze is represented in the Figure 71-03.
In the Figure 71-03 we have a disk of radius R where we highlight a load element
dq located at a distance r from the center of the disk. And this charge element is d from the point P.
To solve this problem we will use polar coordinates. That way, we can write the charge element as
dq = σ r dr dθ. Since the distance of this element to the point P is d, then we can write that d^{2} = r^{2} + z^{2}. With this data we can use Coulomb's law and write
eq. 71-20
Since we want to calculate the electric field at point P located along the z axis, then
we must calculate the component of dE in the direction of the z axis, that is
dE_{z} = dE cos φ.
Therefore, we must write
eq. 71-21
Looking closely at the eq. 71-21 we realize that there are two infinitesimal variables.
So, we must do a double integration: one in relation to r and the other in relation to & theta; .
In addition, we can express cos φ = z/d = z/ (z^{2} + r^{2})^{1/2}.
After an algebraic arrangement, we get the following expression:
eq. 71-22
Note that integration in θ is trivial, as its value is equal to 2 π. Therefore, reorganizing the terms and removing the terms that do not depend on r from the integral, we obtain:
eq. 71-23
The resolution of this integral does not present any difficulties either, as we can substitute variables. Making u = z^{2} + r^{2} and deriving u we get
du = 2 r dr. Then, replacing these values in eq. 71-23 and after an algebraic arrangement,
we obtain
eq. 71-24
Now we have achieved the integration of a polynomial, where we will obtain
I = u^{-3/2 + 1} / (-3/2+1). Performing the calculation we find I = -2 u^{-1/2}.
However, remember that u = z^{2} + r^{2}. Soon making the necessary substitutions and
organizing the terms we find
eq. 71-25
And so, after some simplifications, we find the final result for the electric field produced by an electrically charged disk on its axis of symmetry, or
eq. 71-26
With this result, we can make a prediction of the field value when z → 0.
Note that the second portion, which is enclosed in square brackets, goes to zero when we multiply by
z.
And the first portion is equal to ONE , when we multiply by z. So what we got was the electric field equation on an electrically charged infinite plane. By the eq. 71-27 it is easy to see that the electric field of the infinite plane is constant.
eq. 71-27
On the other hand, it is possible to make a prediction of the electric field value when
z → ∞. For this, let's rewrite the eq. 71-25 and eq. 71-26 as follows:
eq. 71-28
Looking only at what is between the brackets, we can make an expansion by the formula
Taylor. Before we pass the z that is in the numerator into the radical and factoring
we find:
eq. 71-29
Note that the numbers one will be canceled, leaving only the fraction with a positive sign,
because we have the multiplication of two negative signs. Thus, we obtain:
eq. 71-30
See that in the numerator we have the product π R^{2}, which is exactly the
area of the disk. And when we multiply the area by the surface charge density, we obtain the total charge, Q, of the disk. In this way, what
we prove is that when z → ∞ the disk behaves as a punctual charge and due to
this to eq. 71-30 takes the form of Coulomb's law.
Capacitors are of fundamental importance in electrical circuits, both in DC
as in AC. In Figure 71-04 we show in schematic how a capacitor is formed.
Notice that we have two electrodes, facing each other, each having an area, A, and they are
apart from each other, d . In addition, one of the electrodes has a charge + Q
and the other has a charge - Q . This arrangement of two electrodes, charged with the same
absolute charge value, but with opposite signs, is called the capacitor of parallel plates.
It is worth mentioning that the total charge of a capacitor is null , since the capacitor charge occurs through of a means of transferring electrons from one plate to another. Thus, the plate that gains electrons acquires charge - Q = n (- e) , while the one that loses electrons acquires charge + Q .
The idea here is to determine the value of the electric field created on both sides of the plates, that is, between the plates and out of them. Due to the fact that opposite charges attract each other, all charges meet on the internal surfaces of the two plates. Therefore, internal surfaces can be considered
as charged planes with equal and opposite surface charge densities.
Notice in Figure 71-05 , that the electric field created by the plate with positive
charge points out of the surface, while the electric field created by the negative charge points
in the direction corresponding surface.
It is easy to see that the two fields are parallel, have the same direction and sense,
and also have the same intensity.
In a capacitor the distance d is much smaller than the area A of the capacitor plates. It was studied in chapter 3 that the larger the area of the capacitor plates and the smaller the distance between them, the greater the capacitance of the capacitor (C = ε A / d).
Turning our attention to Figure 71-05 , we realized that to determine the resulting field inside the capacitor we can use the principle of superposition. Thus, adding the two fields together, we have the resulting field and it points from the positive to the negative plate. On the other hand, outside the capacitor the fields point in opposite directions, and as we saw earlier, the field of a charge plate is independent of the distance to the plane, so they have the same module. Consequently, fields
cancel each other out outside the capacitor plates.
In this way, it is possible to calculate the field between the capacitor plates starting from the field created by a charged infinite plane, studied in item 4.5 and defined by eq. 71-27 , shown below for clarity. Thus, the field corresponding to the positively charged plate, points in the direction of the negative plate and has a module equal to
And the field produced by the negative plate has the same module as the above equation, and also points from the positive plate to the negative plate, therefore, we can add the two fields and thus obtain the
resulting field between the capacitor plates.
eq. 71-31
where σ is the surface charge density of the plate and A is the surface area of each capacitor plate.
We are interested in analyzing how an electrically charged particle behaves
when it is subjected to an electric field. For our study we will assume that the particle has mass
m and has an electrical charge q. We have already studied in item 2 that, in this case,
the particle is under the action of a force due to the electric field and given by eq. 71-02,
repeated below:
eq. 71-02
This relation between field and force constitutes the definition of an electric field.
Note that the force exerted on a negatively charged particle has the meaning
opposite to the electric field vector. Algebraic signs are important!
This force will cause the charged particle to accelerate with an acceleration given by
eq. 71-32
This acceleration is the charged particle's response to the electric field to which it is subjected. Watch in eq.71-32 , that the q / m ratio, known as charge-to-mass ratio, assumes a character very important in the dynamics of the movement of a charged particle. Thus, two particles with the same charge may experience different accelerations if they have different masses. On the other hand, particles with different charges and masses will experience the same acceleration and follow the same trajectory, if they have the same reason mass-charge . In addition, we can add that a charged particle when moving through an uniform electric field, has numerous practical applications due to its simplicity of movement, perfectly defined by kinematics. Da eq. 71-32 we can say that
"Every electrically charged particle, in the presence of a uniform electric field,
will move with constant acceleration."
The basic trajectory of an electrically charged particle in a uniform electric field is an
parabola , similar to the ballistic displacement of an object in the uniform gravitational field
close to the Earth's surface. It is important to add that to determine the orientation of the acceleration
of the charged particle we must determine the electric field vector. We must be aware of the fact that
if we launch the charged particle parallel to the electric field vector, the movement
will be one-dimensional .
In many devices that use a cathode ray tube, such as televisions, computer monitors, oscilloscopes, etc ..., parallel electrodes are used to deflect charged particles. Figure 71-06 shows an internal diagram of a CRT (cathode ray tube).
Note that we have a filament that, after reaching a certain temperature, releases electrons.
These, in turn, are accelerated by another (called screen ) that has a hole in the center to allow the electron beam to pass. This electron beam reaches a great speed,
reaching 10% of the speed of light. This set of electrodes is called
electron cannon .
There are two more sets of parallel electrodes. One of them is called vertical baffles (shown in
Figure 71-06 ) and the other of horizontal baffles (not shown in
Figure 71-06 ), both in charge of producing a vertical and horizontal displacement of the electron beam.
After leaving the baffle plates, the electron beam moves (through the vacuum,
in order to avoid collisions with air molecules) directly to the cathode ray tube screen,
where the electron collides with a phosphor coating on the inner surface of the screen, producing
there a bright spot. The screen is powered by a source of very high electrical voltage in order to
attract the electron beam. By properly adjusting the electric field values between the baffles,
through the variation of V_{d}, electrons will be directed to any given point on the screen.
Example 71-2
Source:
Example 27.9 - page 837 - KNIGHT, Randall D. -
Book: Física - uma abordagem estratégica - 2ª edição - Ed. Bookman - 2009.
An electron gun generates an electron beam that moves horizontally with speed
of 3.3 x 10^{7} m/s . The electrons travel through an empty space of 2.0 cm
wide between two parallel electrodes, where the electric field is 5.0 x 10 ^{4} N / C
oriented downwards. In what orientation (angle and direction) is the electron beam deflected by these electrodes?
Solution
In the statement of the problem it is said that the electric field has a downward orientation. This, in itself, leads us to conclude that the upper deflector plate is positively charged and,
consequently, the lower deflector plate is negatively charged. So the electron beam, which has a negative charge, will be attracted to the positively charged plate. Then, the electron beam will make
a parabolic curve with an upward orientation.
Note, by Figure 71-07, that the electron beam has an initial horizontal velocity of
v_{oH} = 3.3 x 10^{7} m/s and has no initial vertical speed, that is,
v_{oV} = 0 m/s. Therefore, the electron beam as it passes between the baffles,
will keep the horizontal speed constant, as there are no horizontal forces acting on the beam. However,
this does not happen in the vertical direction, since the deflector plates exert vertical forces on the
electron beam. In this case, the electron beam will acquire an acceleration vertically and, consequently, its final speed, vertically, will be different from zero. So, we must calculate this acceleration and, for that, we will use the eq. 71-32 . Thus, replacing with the numerical values, and remembering that the mass of the electron is equal to m_{e} = 9.11 x 10^{-31} Kg, we find
a_{V} = (1.60 x 10^{-19})(5.0 x 10^{5}) / 9.11 x 10^{-31} =
8.78 x 10^{15} m/s^{2}
As previously stated, the horizontal velocity is constant and, therefore, we can calculate the time it takes the electron to travel the 2.0 cm width of the electrodes. So
t = L / v_{oH} = 2 x 10^{-2} / 3.3 x 10^{7} = 6.06 x 10^{-10} s
As we know the vertical acceleration of the electron beam, we can calculate the vertical speed that the beam will gain when passing through the electrodes at time t , remembering that the initial vertical speed is zero.
v_{oV} = a_{V} t = 5.32 x 10^{6} m/s
We must be aware of the fact that although the beam has gained a vertical speed, it continues with the
horizontal speed, as it is not affected when passing between the electrodes.
So, the final speed is a composition of two speeds: one horizontal and one vertical. Soon,
we can write the final speed in its Cartesian form, as below:
v_{final} = 3.3 x 10^{7} î + 5.32 x 10^{6} ĵ m/s
It is also possible to write the final speed in its polar form. For this we must calculate the module and the beam offset angle. To calculate the module, just use the Pythagoras theorem. And for the angle of deviation we have
θ = tg^{-1} v_{oV} / v_{oH} = +9.16°
So in polar form, we get:
v_{final} = 3.34 x 10^{7} ∠ +9.16° m/s
Note that the positive value of the offset angle already indicates that the beam's offset orientation is upwards. It is also important to understand that after the electron beam passes through the electrodes there are no forces acting on the beam and it follows a straight path from that point, as shown in Figure 71-07.
The law of Gauss and the law of Coulomb are equivalent in the sense that one can
be derived from the other. In practice, Gauss's law makes it possible to determine some static
electric fields that would be very difficult to obtain from Coulomb's law. In turn, Gauss's law
is more general, since it applies not only to electrostatics, but also to the electrodynamics of
fields that change over time. The law of Gauss, to be applied, depends a lot on the
symmetry of the problem in question. Thus, there may be cylindrical,
spherical, etc ... In electrostatics, we can say that a given charge distribution
is symmetric if there is a group of geometric transformations that do not
cause any physical changes.
5.1 Electrical Flow Calculation
In the literature, we often have the electric field strength associated with a given
number of field lines. These lines can come off the charge, if it is positive, or they can be oriented
towards the charge, if it is negative. The electrical flow is associated with the number of field lines that
pass through a certain area, A , of the considered surface.
To understand the concept of electrical flow, let's imagine a conductive loop of rectangular shape
showing an area A = a b , where a and b are the dimensions of the sides of the loop, and that immersed in a uniform electric field, as shown in Figure 71-08.
"Let's define electric flow as the amount of electric field that crosses the effective
area of the loop."
The effective area of the loop is given by A_{ef} = A cos θ = a b cos θ.
We must pay attention to the fact that the angle θ is the angle of inclination of the
axis of the loop in relation to the magnetic field. So, let's define an area vector,
A^{→} = A n^,
with the direction of n^, that is, perpendicular to the loop surface and with modulus equal to the area A of the surface. If the electric field is uniform, then we can write the
electrical flow as the scalar product between the field vector and the area vector.
So this can be written as shown in the eq. 71-33.
eq. 71-33
Note that by the above equation, if the field vector and the area vector are perpendicular to each other, the total flux will be null , as shown in Figure 71-08 . From what has been studied so far, we can draw the following conclusions:
There is an outward flow through a closed surface around a positive net charge.
There is an inward flow through a closed surface around a negative net charge.
There is no resulting flow through a closed surface around a region of space in which the net charge is zero.
In the previous item, our considerations were with respect to an uniform electric field throughout
of a surface. In this item we will analyze the case when the electric field is not constant, that is, it can vary along the surface. One of the ways to calculate the electrical flow on the surface is to divide it into small areas and then calculate the flow in each area. And to determine the total flow, just find the sum of flows for each small area. However, mathematics provides us with a powerful tool to solve this sum, using calculus. For this, we will define a small area by dA. Thus, by making these small areas of infinitesimal size, there will be an infinite amount of them along the total surface. In this way, the summation can be transformed into an integral, and the flow of the electric field across the total surface can be expressed as:
eq. 71-34
The integral that appears in eq. 71-34 is known as surface integral. The law of Gauss establishes a relation between the flow of electric field through a closed surface and the charges that are inside of that surface. Note that if there are no charges inside the considered surface, the electrical flux on the surface is NULL.
To understand how Gauss's law relates the flow of the electric field within a surface
Gaussian with the charge inside that same surface, any surface is chosen
with a charge Q inside, such as the surface shown in Figure 71-09.
We choose a Gaussian surface, such as a sphere of radius r and that involves
fully charge the Q. In this case, the electrical flow through that spherical surface is given
by eq. 71-34 . Assuming Q as a positive charge, we see in Figure 71-09 the field
electrical radially coming out of the surface. The same happens with the vector dA.
Thus, the vector product becomes a product of the modules. Since |E| is constant
on the sphere surface, it is possible to remove it from the integral. And, of course, that the integral
surface area of a sphere is the spherical surface area, equal to 4 π r^{2}.
Therefore, we can write the electrical flow as:
eq. 71-35
Making the necessary simplifications in eq. 71-35, we can write Gauss's law in integral form, for a closed surface, such as:
eq. 71-36
The eq. 71-36 is valid even if there is more than one charge inside the considered surface, because in that case, we can use the superposition principle. Thus, the value of q_{int} will be equal to the sum of the
internal surface charges.
To express Gauss's law in its differential form, we must apply the divergent to the field and explain the internal charge as a function of the volumetric charge density, ρ . Then, applying the Stokes theorem we get the differential form, as shown in eq. 71-37.