Electric field

Vector electric field

Electrostatics study the force of two Q, Q’ point electrical charges between them; if we look only at one of the charges, say the Q, we realize that if we place any Q’ load in any position of space surrounding Q, a force will be exerted on Q’, it is as if the space surrounding Q has a new property, that of exerting a force on any Q’ load in any position that we place it.

Let’s place a small known value q test load at any position in the space given by the position vector v; if we see that an F-force (a vector) is exerted on the charge simply because it is situated at that point, we infer that space is affected by some other unknown Q load that creates that effect, and then the law of Coulomb will be fulfilled,

F-R-qQ/r3[1]

where r is the position vector of q relative to Q. Figure 1 shows the loads q, Q, their position vectors v, w, the relative position vector of q to Q, r s v – w, and the resulting force F, in the direction of r, which is assumed repulsive (loads are of the same sign and repel each other).

Fig.1: electrostatic force between two charges

Assuming the view that the space around Q is affected by that load, since a force is exerted at a distance on any load q at any position v, we can define a vector E such that the force exerted on any load q is simply F s qE from the expression [1]:

E s k(r/r3) Q[2]

This vector is called the electric field vector, and does not depend on the test load q, only on the Q-causing load. If we take two test loads q, q’, and measure the vectors force F, F’, looking for the intersection of the support lines of the forces we will find the origin of the field: the position of the Q load that generates it (Figure 2). In addition, the magnitude of the force F, known the q load, also determines the Q load through the equation,[1] and therefore determines the E field.

Fig.2: two test loads q, q’ determine the origin of the electric field, the Q load, and the vector field E

The property that all the support lines of the electrostatic forces are cut at one point can be expressed by saying that the electric field is a central field because all forces start from a central point of space.

In Physics the concept field is used to describe how a measurable physical magnitude is “distributed” by space; so, we can talk about electric fields, magnetic fields, and even velocity fields in a fluid. If the magnitude is scalar, the field will be, if vector, the field is vector, and if the magnitude is a tensor, the field will be tensory. Specific mathematics for describing fields has been studied in depth, giving rise to the branch of Mathematical Physics known as Field Theory.

Electric field produced by several point loads

Fig.3: Single-load single-charge electric field lines are straight cut into the load

In the case of a single point charge we have seen that all the straight supports of the forces, straight that we will call lines of force of the field or simply lines of the field start from a central point where the single load is located (fig. 3).

What happens to lines if the field is generated by two loads? Since the electrostatic force is cumulative (the contributions of all loads are added) the electric field E will be, and at each point in space the field vectors corresponding to each load will be added. In addition, the force F and field E decrease with the square of the distance, so the lines of the field curve; a field line passes through each point in space, so that the electric field at that point is tangent to the line. Figure 4 represents two equal positive loads, and a small test load in which the two vectors E, E’ generated by the field sources are added. The resulting vectors do not point to any center, the resulting field is no longer central, and the field lines (in black in the figure) are curved.

Fig. 4: field and field lines generated by two equal point loads

It should be borne in mind that any load will generate its own electric field that will overlap existing ones; that is why we are talking about placing “test burdens” which are supposed to be much smaller than the field-generating loads, so that their contribution can be disregarded. In addition, source loads, even though they are much larger than test loads, are supposed to be of point dimensions to avoid mathematical complications.

Electric dipole field

An important particular case is that of the so-called electric dipole, which are two loads of equal magnitude q and counter sign separated by a small d distance (fig. 5).

Fig. 5: field lines produced by an electric dipole, source: By Geek3 , from [GFDL (http://www.gnu.org/copyleft/fdl.html) or CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0)]Wikimedia CommonsIt def

ines the electrical moment of the dipole, p, by the vector

p s qd,[3]

being d the vector that starts from the negative charge and ends up in the positive charge. The distance d is supposed to be much smaller than the distances at which we will place the test loads where we will measure field E, this allows to simplify its mathematical expression, which is obtained from adding the contributions of each load to the total field, turning out to be:

\boldsymbol E=qk\left(3\frac{d\cos\left(\theta\right)}{r^4}\boldsymbol r-\frac{\boldsymbol d}{r^3}\right) [4]

where r is the vector that starts from the center point of the dipole (between its two charges); alternatively, using[3] :

\boldsymbol E=k\left(3p\frac{d\cos\left(\theta\right)}{r^4}\boldsymbol r-\frac{\boldsymbol p}{r^3}\right) [5]


Example 1: We place a load of +10⁻3C at the coordinate origin, and another load of -10⁻3C at the point (1, 0, 0). Calculate the electric field at point P(0.5, 1, 1). If we place at that point a small load of +10⁻⁶C, what force will the field exert on it?

The E’ field due to the first load will be, applying [2]:

\begin{array}{l}E'=k\cdot\lbrack{(0.5,1,1)-(0,0,0)}/\sqrt{(0.5²+1²+1²)\rbrack}^3\cdot(10⁻³)=\\\;(0.5,\;1,\;1)\cdot\frac{10⁻³k}{\left(3/2\right)^3}\end{array}

For the second load:

\begin{array}{l}E''=k\cdot\lbrack{(0.5,1,1)-(1,0,0)}/\sqrt{(0.5²+1²+1²)\rbrack}^3\cdot(-10⁻³)=\\-\;(-0.5,\;1,\;1)\cdot\frac{10⁻³k}{\left(3/2\right)^3}\end{array}

The total field will be the sum of the above:

E=10^{-3}k\left(\frac23\right)^3\left(1,0,0\right).

The force exerted on the test load q is given by F s qE:

\boldsymbol F=q\boldsymbol E=10^{-6}\cdot10^{-3}\cdot k\left(\frac23\right)^3\left(1,0,0\right)=9\cdot\cancel{10^9}\cdot\bcancel{10^{-9}}\frac8{27}\left(1,0,0\right)=\left(1,0,0\right)

a force of 1 Newton in the direction of the X axis.


Electric field created by a load distribution

When instead of point loads we have loaded material bodies we model them as if they contained point loads distributed throughout the body, so that the electric field produced by the body is obtained by adding the contributions of the point loads; Depending on the mathematical form we give to the distribution, the sum can be more or less direct, simple, or complicated. In the limit case, which is in fact the usual, in which we consider that there are a myriad of point loads it will be necessary to use differential calculation and integral.

Electric field created by a flat and homogeneous load distribution

The simplest case of distribution of countless point loads is that of a thin bar of 2L length that has loads only on one side and are also evenly distributed; let’s call the electrical charge density per unit of length, which will obviously be Q/2L being Q the total load of the bar. and let’s just calculate the electric field at a point P above the bisector of the bar, at a height h:

Fig. 6: Geometry for the calculation of field E produced by a homogeneous load line at a point above the bisector of the line

Taking a differential element of length dx and located at a distance x from the center of the bar, by the geometry of the problem we see that the distance r2 will be equal to x2 + h2, and the differential load of that element will be dQ s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

\operatorname dE_1=k\frac{dQ}{r^2}=k\frac{\rho\cdot dx}{x^2+h^2}

Given the symmetrical situation point P with respect to the bar, there will be another dx element located in -x that will produce a differential field dE_2 such that when you sum the vectors dE_1 with dE_2 the horizontal components will be overridden and the resulting will be vertical, dE with a module:

\operatorname dE=\operatorname dE_1\sin\left(\theta\right)+\operatorname dE_2\sin\left(\theta\right)=2\cdot k\frac{\rho\cdot dx}{x^2+h^2}\cdot\sin\left(\theta\right)

This will happen along the entire bar, so we conclude that the resulting field must be vertical. We now have to add all the differential contributions along the bar:

E=\int_0^L\operatorname dE=2k\int_0^L\frac{\rho\cdot dx}{x^2+h^2}\cdot\sin\left(\theta\right)

Note that the integration limits are and a[0, L]re not[-L, L] for each dx element to the right of the bar (in x) we have already added the contribution of the symmetrical element to the left (in -x). Considering that -sin-left(-theta-right)-frac hr-frac h-sqrt-x-2+h-2- we reach the integral

E=2k\int_0^L\frac{\rho\cdot dx}{x^2+h^2}\cdot\frac h{\sqrt{x^2+h^2}}=2kh\rho\int_0^L\frac{dx}{\left(x^2+h^2\right)^{3/2}}

Without going into details of the integral calculation (we can calculate it for example using WolframAlpha), we will have:

E=2kh\rho\frac L{h^2\sqrt{h^2+L^2}}=\frac{2k\rho L}{h\sqrt{h^2+L^2}}

In the event that the bar is very long compared to the distance h, i.e. L >> h, we can simplify the value of the field:

\lim_{L\rightarrow\infty}E=\frac{2k\rho}h\lim_{L\rightarrow\infty}\frac L{\sqrt{h^2+L^2}}=\frac{2k\rho}h [6]

The value of the field will vary depending on 1/h.


Vector field flow. Solid Angle

In the previous section we have seen that even in a simple case (one-dimensional load line, homogeneous, field at a point of the bisector…) when calculating fields due to continuous distributions of loads immediately appear complicated integrals, or very complicated. In this section and the next we see a view of the geomeath-based issue that often greatly simplifies calculations,

Consider a central vector field, which is the one that makes each P point of the space correspond to a vector V that follows the OP direction, being O the center of the field, a fixed point (Figure 7).

Fig.7: Central field, the agent mass located in O

The special case of the form V = r·c/r² where r is the OP module (the distance to the center of the field), r the unit vector indicating the direction of OP, and c is a constant that we call the field’s agent mass; it’s called the Newtonian field. They are Newtonian fields, the electric field, the magnetic field and the gravitational field. Let’s define the flow of the vector field through an infinitesimal dS surface, which we calld\phi, as the scalar product V·dS, where dS is the vector perpendicular to the surface; being infinitesimal dS (very small), we will consider its curvature to be negligible, and therefore flat (Figure 8):

Fig. 8: surface differential element, vector dS, and vector field V that we assume passes through the center of the surface

\operatorname d\phi=V\cdot dS=V\cdot dS\cdot\cos\left(\theta\right) [7]

Next, let’s define the solid angle of the dS surface relative to the origin of the center field O. Let’s join the ends of the dS surface with the O point using OP lines, and define a C sphere of radius 1 with center in O; the OP lines will cut to the sphere by defining on it a small surface of dS over C (Figure 9).

Fig. 9: solid angle subtended on sphere C by surface element dS

Now let’s look at a geometric property of Newtonian fields: the flow of the field through dS is:

\operatorname d\phi=V\cdot dS=V\cdot dS\cdot\cos\left(\theta\right)=\frac c{r^2}dS\cdot\cos\left(\theta\right)=\frac c{r^2}dS' [8]

where we’ve matched dS\cdot\cos\left(\theta\right)=dS', which is the projection of the dS surface over the perpendicular to the field vector V. The surfaces dS’ and dΩ (Fig. 10) are parallel, and are joined by the same lines to the center point O, then it is true that the reason for their areas is equal to the reason for their distances to center O squared:

Fig. 10: dS’ and d’ are parallel, and are attached by the same lines to the central point O

\frac{\operatorname d\phi}{dS'}=\frac1{r^2}\Leftrightarrow dS'=r^2\operatorname d\phi

If the reader don’t see why, think that the area of the sphere is 4πr², the area of the unit sphere is 4π, and the area of a sphere that passes by dS’ is 4πr², so the ratio of areas is 4πr² : 4π = r². Using this ratio in the equatio[8]n we get:

\operatorname d\phi=\frac c{r^2}r^2\cdot d\Omega=c\cdot d\Omega [9]

which tells us that the flow of the Newtonian field vector across any differential surface does not depend on the distance r, and is directly equal to the product of the agent mass of the field by the solid angle subtended by the surface over the unit sphere centered on the origin of the field.

Flow of the field through closed surfaces

The scalar product defined in[7] can be positive or negative depending on the relative orientation of the field vectors V and dS; consider a closed surface S, let’s calculate the total flow of the field through S integrating for each dS element:

\phi=\int_S\operatorname d\phi=\int_SV\cdot dS  [10]

If the total flow is positive, we will say that it is an incoming flow in S, and if it is negative, it will be an outgoing flow of S. If the center of field O is on the outside of S, the OP lines from the center passing through S will cut to S by an even number of points; however, if O is inside S, the lines will cut to S by an odd number of points (Figure 11).

Fig.11: flows from the O center through closed surfaces

Therefore in the outer case each line will create a succession of inbound-outbound-inbound-outgoing-etc flows in even number, that is, of alternating positive and negative flows; we have seen that the flow does not depend on the distance to the center (equation [9]) but only on the solid angle subtended, which will be the same for each surface element over the lines starting from O. Therefore, for an outer O point, inbound and outbound flows have the same value and override each other, resulting in a total null flow:

In a Newtonian field, the total flow through a closed surface that does not contain the center of the field is null.

However, if the surface contains the center of the field we will have an odd number of incoming-outgoing flows, and their sum will not be overridden; in fact its value is given by the theorem of Gauss, which is derived from the equations [9]and[10] :

\phi=\int_S\operatorname d\phi=\int_Sc\cdot\operatorname d\Omega=c\cdot\int_S\operatorname d\Omega=4\pi c  [11]

Expressed in words:

In a Newtonian field, the total flow through a closed surface containing the center of the field is equal to 4\pi multiplied by the value of the agent mass.

In the case of the electric field, the agent mass is worth k-Q, Q being the electrical charge.

If we have a set of loads, each load will create its field flow, and the total flow will be the sum of all of them.

Field created by an infinite distribution of loads, flat and homogeneous, using the Gaussian theorem

As an example of the usefulness of the concept of field flow and theoriema of Gauss will calculate the electric field created by a uniformly loaded plate, which we will assume to be very large, over a P point located at a height h of the plane.  Imagine another P’ point located on the other side of the plane, symmetrical to P, and think of a cylinder with PP’ axis and upper area dS (in Figure 12 we see a side view of the loaded plane and the situation).

Fig. 12: Using gauss’s theorem to calculate electric field E

By symmetry the E field must be vertical and in the directions indicated in the figure; the normal dS vectors to the cylinder bases will have the same direction as E, then the resulting scalar product, taking into account that there will be no flow of E on the sidewalls of the cylinder because it is parallel to the field (then the vector normal to the walls is perpendicular to E and its scalar product, null) is \operatorname d\phi=E\cdot dS+E\cdot dS=2E\cdot dS.  If we call sigma the load density per surface unit of the board, the load enclosed inside the cylinder will be \operatorname dQ=\sigma\cdot\operatorname dS. Because of the Gaussian theorem, the flow must then be \operatorname d\phi=4\mathrm\pi\cdot\mathrm k\cdot\mathrm\sigma\cdot\operatorname d\mathrm S; matching the two expressions for the flow we get:

d\phi=4\pi k\sigma\cdot\cancel{dS}=2E\cdot\cancel{dS}\Leftrightarrow\boxed{E=2\pi k\sigma}

We see that the ingensity of field E created by a homogeneously charged infinite plate does not depend on the distance h to the plate, a remarkable result.

Problems

  1. Calculate the electric field produced by a sphere loaded with homogeneous load density, both inside the sphere and outside.
  2. A sphere loaded with homogen load density has a spherical cavity inside it, the centers of the loaded sphere and the cavity are at a distance d. Calculate the electric field in the cavity.


Solutions

Problem 1 – In the figure we see the geometry of the problem: we represent a spherical surface S inside and concentric to the loaded sphere (in blue) and on S any point P, by which we draw a line that passes through the center O and divide the spheres into two symmetrical halves; by symmetry, the vector field E(P) at point P cannot be directed towards either of the two halves in particular, so it must be radial. In addition, point P could be any point located in S because we have spherical symmetry, then the module of field E will be the same throughout S, that is, the value of E depends only on the radius of S.

By symmetry of the problem with any line that passes through O, the E field must have the same module on every inner S sphere, and it must be radial

Let’s apply gauss’s theorem to surface S: first of all we give mathematical form to the above considerations about field E:

\overrightarrow E(P)=\frac1rE(r)\cdot\overrightarrow{OP},

that is the vector field E is equal to the E(r) module by the radial vector OP divided by the OP module, which is r. We can define the radial unit vector , and the expression of the field at all point P of S is more compact: \overrightarrow E(P)=E(r)\cdot\widehat r.

Now we calculate the flow of E through the S surface, applying [10]:

\phi=\int_S\operatorname d\phi=\int_S\overrightarrow E\cdot d\overrightarrow S=\int_SE\left(r\right)\widehat r\cdot d\overrightarrow S=E\left(r\right)\int_SdS=E\left(r\right)\cdot4\pi r^2

where we have applied that the vector differential element of surface dS is a radial vector of dS module, and therefore its scalar product with the unit radial vector is simply dS, in addition E(r) is constant over S, then it can exit outside the integral, and this integral over S of the dS element is simply the surface of the sphere S. This flow that we have calculated, according to the Gaussian theorem [11], must be equal to \phi=4\pi kQ, where Q is the load contained in the inner volume to S; calling \rho to the load density per unit volume, we have:

Q=\int_V\rho\cdot\operatorname dV=\frac43\rho\pi r^3\Rightarrow\phi=4\pi k\cdot\frac43\rho\pi r^3=\frac{16}3k\rho\pi^2r^3 [12]

By matching this flow given by the Gaussian theorem with which we calculated earlier, we find the module of field E:

E(r)\cdot4\pi r^2=\frac{16}3k\rho\pi^2r^3\Rightarrow\boxed{E(r)=\frac43k\rho\pi r} [13]

We see that the E(r) dependency is linear: it increases linearly with r. In the international system of units the constant k is expressed in function of the so-called electrical permittivity of the vacuum -v</em>arepsilon_0, and the field is reduced to

E(r)=\frac43\frac1{4\pi\varepsilon_0}\rho\pi r=\frac1{3\varepsilon_0}\rho r. [14]

This expression is valid for r\leq a, being a t the radius of the loaded sphere. For distances r to the center of the sphere that are greater than the radius to the calculation is very similar, only that the value of the Q load is constant, being the total load of the sphere, Q-frac43-pi a-3-rho, and when replacing it in the Gaussian theorem, the total flow across a radio surface r > to -phi-frac43-pi a-3-rho-cdot4-pi k , matching this flow with the vectorly calculated:

'phi'frac43's'4'p'r'2'Leftrightarrow's 'left's's'left')right's'right's''frac's'left')'s'3'''2''2'''leftrightarrow''[14b] which, based on-varepsilon_0will be worth E-left(r-right)-frac43-pi-frac-a-3-r-2-rho-frac1-4-pi-varepsilon_0-frac-rho-a-3-3-varepsilon_0r-.. [15]

So inside the sphere the E field grows linearly with the distance to center, and on the outside it decreases squarely:

Variation of field E with distance r to center in the case of a radio sphere to evenly loaded

Problem 2 – The complication of this problem is to see how to treat the cavity inside the sphere; the easiest way is to realize that the situation is equivalent, in terms of the calculation of the E field, to assume that the sphere, from radius R and center to O, is evenly loaded in its entirety with a load density of -rho, and that the cavity is another spherical surface, radio r< R y centro O’, a la que añadimos, superponiéndola, otra densidad de carga de igual valor pero signo contrario, -rho, con ello, la carga neta en la cavidad será cero. So, because the electric field is cumulative, we can calculate the field E by overlapping the field created by the entire radius sphere R, which we call E_O, plus the field created by the negative load in the radius sphere r, which we call E'_O. The geometry of this scheme is seen in the image, where we have drawn a P point either within the cavity, the two fields generated at that point, and the total field.

Vectors E and geometry of the problem

To calculate each field we apply t[13]he P point as it is interior to the two spheres considered:

'overrightarrow E'left(P'right)'overrightarrow E'_O'left(P'right)+'overrightarrow E'_O'left(P'right)'frac43k's "cdot"widehat r-'frac43k''ah'rho r''cdot'widehat r'''frac43k''p'2'left(r'cdot'widehat r-r''cdot'widehat r''right)

We need to r-cdot-widehat r-overrightarrow r,-;r'&r'&gt'>cdot'widehat r''overrightarrow r' and looking at the figure we deduce that r'cdot'widehat r' -r'-cdot-widehat r-overrightarrow r--overrightarrow r'-overrightarrow-OP--overrightarrow-O'P-overrightarrow-OO-'.  Therefore we have the following expression for the field inside the cavity:

'overrightarrow E'left(P'right)'frac43k''th'cdot'overrightarrow'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''frac43k''-th-rho-cdot-overrightarrow d

Note that E is a constant vector directed according to the OO’ line (the module vector d equal to the distance d between centers).

The field inside a cavity within a uniformly charged sphere is constant

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