# 6. Electric Flux

## Introduction

The explanation of the force between particles through the existence of a field comes from the time when the theory of universal gravitation was developed. The difficulty in accepting that a particle can affect other distant particles, without any contact among them, was surpassed in classical physics with the concept of the force field. In the case of the electrostatic force, the mediator field that transmits the electrostatic force was called ether; light would be a wave propagating in ether. In the nineteenth century numerous experiments were conducted to detect the presence of the ether, without any success. By the turn of the century, physicists came to the conclusion that such ether did not exist. However, the electric field has physical existence, in the sense that it carries energy and it may persist even after the charges that produced it have dissapeared. In quantum physics, the electric interaction is explained as an exchange of particles among charges; those exchanged particles are the photons which are also the particles in a light ray. Each charge emits some photons that are absorbed by other charges. In this chapter we use the classical theory of eletromagnetism, in which the electric field is pictured as an invisible fluid that drags the electric charges.

## 6.1. Electric field produced by point charges

The electric field produced by a point charge has been studied in Chapter 1. Figure  1.6 shows the repulsive field produced by a positive charge.

Equation  1.5 for the module of the field produced by a point charge can be written in vector form. If the charge $q$ is at the origin of the coordinates system, the result is:

(6.1)

where $r$ is the distance to the origin and $\hat{r}$ the unit vector in the radial direction away from the charge. If the charge is negative, the above equation is still valid and the vector $\vec{E}$ points in the opposite sense of $\hat{r}$(Attractive field). The unit vector$\hat{r}$ obtained by dividing the position vector $\vec{r}$ by your module, $r$. If the charge is not on the rise but at a position$\vec{r}_1$, Equation  6.1 can be easily generalized, obtaining:

(6.2)

The field produced by a system of point charges is obtained by vectorially adding the fields produced by each of the charges. For example , the left side in Figure  6.1 shows the fields produced at some points for each of two point charges 4 and 9 nC nC. The right side shows the resulting field in the presence of the two charges, which corresponds to the vector sum of the two fields.

Equation  6.2 can be generalized for a system$n$point charges. In Cartesian coordinates, the charges$q_1$, $q_2$, ..., $q_n$ are in points ($x_1$, $y_1$), ($x_2$, $y_2$), ..., ($x_n$, $y_n$) The plane O$xy$ (Generalization for space The$xyz$ should be obvious), the result is:

(6.3)
(6.4)

### Example 6.1

Plot, on the plane $xy$, the lines of the electric field produced by two point charges of 4 nC and 9 nC, placed 1 cm apart from each other.

Resolution . First, one can choose a convenient system of units; measuring the charges in nC and distances cm, the value of the Coulomb constant will be:

Thus, if the value $k$= 90 is used, field units are μN / nC. Asuming that the 4 nC charge is at the origin and the 9 nC charge is at point (1, 0) and substituting in equations  6.3 and 6.4 one obtaines:

This field has been represented at some points in Figure  6.1 . To plot the electric field lines, you can use the program plotdf the Maxima .

(%i1) Ex: 90*4*x/(x^2+y^2)^(3/2) + 90*9*(x-1)/((x-1)^2+y^2)^(3/2);

(%o1)   $\dfrac{360\,x}{\left(y^2+x^2\right)^{3/2}}+\dfrac{810\,\left(x-1\right)}{\left(y^2+\left(x-1\right)^2\right)^{3/2}}$
(%i2) Ey: 90*4*y/(x^2+y^2)^(3/2) + 90*9*y/((x-1)^2+y^2)^(3/2);

(%o2)   $\dfrac{360\,y}{\left(y^2+x^2\right)^{3/2}}+\dfrac{810\,y}{\left(y^2+\left(x-1\right)^2\right)^{3/2}}$
(%i3) plotdf ([Ex, Ey], [x, -1, 3], [y, -2, 2], [vectors,"blank"])$Following this last command you need to click a few points for the field lines that appear in Figure 6.2 . Note that the expression obtained for$E_y$ It is equal to zero in any point of the axis of $x$ ($y$= 0), except for points where the charges. equating$y$ = 0 and $E_x$ = 0 we obtain the coordinates of point where the field is null, $x$ = 0.4 and $y$= 0, which is a saddle point as shown in Figure 6.2 . The points where the two positive charges are we repulsive. ## 6.2. Properties of electric field lines The electric field can be represented by vectors that indicate the value of the field at each point of space, as was done in Figure 6.1 . The problem is that the field varies rapidly with distance, which makes the vectors are very large in some places and too little in others. Representation by field lines is more convenient. The field lines follow the field direction and at every point of these lines, the field is tangential to the line and in the direction indicated by the arrows. The electric field lines have several properties: • In the vicinity of a positive point charge there lines leaving in all directions and near a negative charge there lines entering in every direction (see figure 6.3 ). • Two field lines never intersect; an intersection point of the field would have two different directions, which is not possible. • In the next chapter it is shown that the corresponding Jacobian to the electric field is always symmetric. This implies that the eigenvalues ​​of this matrix are always real and never complex. Thus, the only equilibrium points that may exist in an electric field are nodes and saddle points. A node can be attractive or repulsive; attractive if there is a point where a negative point charge and if repellant is a point where there is a positive point charge. The saddle points are points where the field is zero, with no charge at this point. In the example shown in Figure 6.2 , there is a saddle point (0.4, 0), where the field is zero. There are two field lines that terminate at this point saddle and two field lines therein start. Another example is the field lines of a dipole electric formed by two equal charges of opposite sign. Assuming that the two charges are located in points (-1, 0) and (1, 0) can be drawn field lines with the following commands: (%i4) Ex: (x+1)/((x+1)^2+y^2)^(3/2) - (x-1)/((x-1)^2+y^2)^(3/2)$
(%i5) Ey: y/((x+1)^2+y^2)^(3/2) - y/((x-1)^2+y^2)^(3/2)$(%i6) plotdf ([Ex, Ey], [x, -3, 3], [y, -3, 3], [vectors,"blank"])$

The result is shown in Figure  6.4 .

A continuous charge distribution can be approximated by a series of point charges. For example, if there are evenly distributed charges on the shaft segment $x$ in between $x$ = -3 And $x$ = 3, one can imagine a system of point, equidistant charges on the segment between $x$ = -3 And $x$ = 3. Maxima commands to trace the field lines of these seven specific charges are:

(%i7) Ex: sum ((x-i)/((x-i)^2+y^2)^(3/2), i, -3, 3)$(%i8) Ey: sum (y/((x-i)^2+y^2)^(3/2), i, -3, 3)$
(%i9) plotdf ([Ex, Ey], [x,-20,20], [y,-20,20], [vectors,"blank"])\$

and the obtained graph is shown in Figure  6.5 .

## 6.3. Flux

The electric field may be better understood using the concept of flux. It defines the electric flux by analogy with an incompressible fluid. In the flux of the fluid, the field lines are tangential fluid velocity at each point and the velocity field of the flux is equal to the volume of fluid passing through the surface per unit time.

Through a surface area $A$Perpendicular to the fluid velocity and the velocity module, $v$, Is constant at every point of this surface, the volume of fluid passing through the surface per unit time is equal to $v\,A$. Figure  6.6 shows two examples of the field lines , and in each case one surface perpendicular to the field lines.

By analogy with the fluid flux, where an area of ​​the surface $A$Perpendicular to the lines of electric field, as in Fig  6.6 , if the module$E$the field is constant on this surface, it defines the flux electricity through the surface equal to the product of the times field module surface area:

(6.5)

The volume enclosed by the field lines passing through a closed curve, for example, the boundary of the surface S 1 in Figure  6.7 , is called pipe flux . Using the analogy with the incompressible fluid, if there are no points within the pipe where fluid enters or exits, then the flux is the same in all cross sections of the tube, regardless of the curvature or slope of these sections. For example, in the flux tube in Figure  6.7 , the volume of fluid passing through the three surfaces S 1 , S 2 and S 3 per unit time must be the same and, as such, the flux through these surfaces, or any other section of the tube is equal.

This property of the flux tubes can be used to calculate the flux through a surface that is not perpendicular to the field lines. If the field lines are not perpendicular to the surface but are inclined at an angle$\theta$ in relation to the unit vector $\hat{n}$normal to the surface, as shown in Figure  6.8 , the flux through surface area$\Delta A$It is equal to the flux through this area in the projection plane perpendicular to the field lines, i.e. through surface dashed in Figure  6.8 . This is because the inclined surface and the broken surface part of the same flux tube, formed by the field lines that pass through the border of the two.

The area of ​​the hatched surface is $\Delta A\cos\,\theta$,at where $\theta$It is the angle between the field and perpendicular to the plane (Figure  6.8 ). As the surface dashed but is perpendicular to the field lines, the flux through it is equal to the field modulus times the area, or

(6.6)

Figure  6.9 shows three possible fields on the surface. The field$\vec{E}_1$makes an acute angle with the normal unit vector and, therefore, produces positive flux, ie flux passing in the same direction of the normal unit vector. The field$\vec{E}_2$It is perpendicular to the surface and, as such, its scalar product with the normal unit vector is zero and this field does not produce any flux. Finally, field$\vec{E}_3$ makes an obtuse angle with the normal unit vector, thus producing negative flux, ie flux in the opposite direction of the normal unit vector.

The scalar product $\vec{E}\cdot\hat{n}$It is the field component in the direction normal to the surface. That is, the electric flux is the normal component of the field times the surface area.

In the case of non - uniform fields and curved surfaces, the surface is divided into several segments, as shown  6.10 . If the number of segments is high and each is sufficiently small, they can be approximated by small planes.

in each small plane number $i$ field exists $\vec{E}_i$Approximately constant, and therefore, the flux $\varPhi_i$this small plane is given by equation  6.6 . The total flux through the surface is equal to the sum of the fluxes in all the small planes.

(6.7)

The approximation becomes exact in the limit as $n$and this limit is infinite sum in equation  6.7 is called the integral of the electric field surface, the surface S, and is written as follows:

(6.8)

In cases studied in this chapter this surface integral can be calculated easily, without the need to use the whole surface calculation techniques.

## 6.4. Gauss's law

The electric field produced by a charge distribution is the superposition of the fields produced by many point charges. It should then analyze the electric flux produced by the field of a single point charge$q$. In relation to a surface S closed, the charge$q$ can be either inside or outside of this surface.

If the point charge $q$is within the surface, such as on the left side of figure  6.11 , the flux is exactly equal to the flux through of a sphere centered at the charge, and the right side of the figure  6:11 , because the two surfaces are part of a flux tube; all the field lines that pass through the closed surface also pass through the ball.

The spherical surface, the field is perpendicular and its module is always equal to $E=k\,|q|/R^2$, at where , where $R$ is the radius of the sphere. Thus, the flux through the sphere can be obtained using equation  6.5. The area of the sphere is $4\,\pi\,R^2$ and multiplying the field module, obtains flux:

(6.9)

That is that the total flux produced by the point charge $q$Through any closed surface on which the charge is on the inside, it is always $4\,\pi\,k\,|q|$Regardless of the size of the surface. In the case of closed surfaces, it is customary to always calculate the flux from the surface. Thus, eliminating the absolute value of the charge and writing the stream as$4\,\pi\,k\,q$The sign of the charge will give the correct sign for the flux: if the charge is positive, there is flux going out of the surface, but if the charge is negative there is flux entering the surface.

If the charge $q$is outside S, the surface tangent to the field lines define a flux pipe. The surface S is divided into two parts S 1 and S 2 on both sides of the curve through which the field lines tangent to S (see Figure  6.12 ). The electric flux through the two surfaces S1 and S2 are equal in absolute value, because they are part of the same flux tube, but have opposite signs, since one of the unit vectors$\hat{n}_1$ or $\hat{n}_2$points in the direction of the field lines and the other in the opposite direction. That is, if $q$> 0, there is flux entering S2 and the same flux goes out of S1; if$q$<0, there is flux entering S1 and the same flux comes out of S2. We conclude that the electric flux due to a point charge$q$ It is void in any closed surface, if the charge is off the surface and equal to $4\,\pi\,k\,q$ if the charge is within the surface.

A charge distribution can be divided into multiple point charges $q_1$, $q_2$, ..., $q_n$And the total flux through a closed surface S is equal to the sum of $n$ fluxes produced by each of the point charges. The charges outside S do not produce any fluxe, while each charge $q_i$ inside S produces flux $4\,\pi k q_i$. Therefore, the total flux through the closed surface S is:

(6.10)

at where $q_\mathrm{int}$is the total charge inside the surface S. This equation is called Gauss law:

The electric flux through any closed surface is equal to the value of the total charge inside the surface, multiplied by $4\,\pi\,k$.

If the total charge inside is positive, the flux is positive, indicating that there are field lines come out to the surface. If the total charge is negative, the flux is negative because there field lines entering the surface.

The total electric flux around a point charge is directly proportional to the charge value. In some cases it is possible to draw a number proportional to the charge field lines to give a more approximate idea of the amount of flux in different regions; For example, in Figure  6.2 were drawn 8 field lines to leave the charge nC 4, lines 18 and to leave the charge of 9 nC.

Gauss's law is very useful for calculating electric fields of systems with symmetry, as discussed in the following sections.

### Example 6.2

An electron is in the center of a cube whose edges measures $20\;\mathrm{nm}$. Calculate the magnitude of the flux through one of the cube faces.

Resolution . The total surface of the hub is closed and thus the flux through it can be calculated easily using Gauss's law:

The internal charge $q_\text{int}$It is the charge of the electron (-1602 × 10 -19  C). Thus, the total flux through the cube surface is:

By symmetry, the flux through each face must be the same, so that the flux on one side is the sixth part of the total flux in the cube: -3.02 nN · m 2 / C. The negative sign of the result indicates that the flux is into the hub.

### 6.4.1. Field of a plane

Consider a plane, with charge evenly distributed. Seen in section, the plane appears as a line segment, and the field lines are similar to lines represented on the right side of the figure  6.4 .

In regions near the central plane, the field lines are approximately parallel to each other. The higher the level, the greater the area where the lines are approximately parallel. In the idealized case of an infinite plane, even if the lines are parallel and the field's value depends only on the distance to the plane, since the plane of the appearance is the same at any point.

To calculate the electric field using the Gauss' law, imagine a cylinder with the base and top parallel to the plane, as shown in Figure 6.13 .

The sides of the cylinder walls are not crossed by the electric flux, because the field is parallel to the surface. In each of the circular covers the cylinder, the field is perpendicular and has constant module, for all points on the cover are the same distance from the plane. The flux in the top and bottom of the cylinder is then$A\,E$, on what $A$ is the area of ​​the cylinder base and the total flux through the cylinder surface is:

(6:11)

According to Gauss's law, the flux is also equal to:

(6:12)

at where $Q$is the charge on the plane that is inside the cylinder. Equating the last two equations gives the field module:

(6:13)

on what $\sigma$is the charge surface , i.e., charge per unit area:$\sigma = Q/A$.

### 6.4.2. Field of a straight wire

Consider a very long straight wire, with charge evenly distributed. The field lines should have the radial direction. Imagining a closed surface that is a cylinder radius$R$ and then $L$With axis of the wire, as shown in Figure 6.14 it follows that the flux is zero and circular covers of the cylinder, since on them the field is parallel to the surface; the side wall of the cylinder, the field is perpendicular and constant modulus. Thus, the total flux will be:

(6:14)

at where $E$ It is the fi eld distance $R$wire. According to Gauss's law, this flux should be also equal to

(6:15)

at where $Q$ is the charge wire which is inside the cylinder S. equating the two equations above, one obtains the field module:

(6:16)

at where $\lambda$It is the charge linear (charge per unit length):$\lambda = Q/L$.

### 6.4.3. Field of a conducting sphere

A conductive ball, with charge $Q$ and radius $a$, The repulsive force between charges of the same sign, causes the charge to distribute uniformly on the surface of the sphere. then there is spherical symmetry and the field lines should point in the radial direction.

To calculate the field, imagine a sphere of radius $r$Concentric with the conductive ball. The field is perpendicular to the surface and has constant modulus$E$; as such, the value of the electric flux is:

(6:17)

Since the charge is distributed on the surface of the conductive sphere, are within the no charge and, according to Gauss's law, the flux through the sphere of radius $r$ is void if $r$ < $a$Or equal to $4\,\pi\,k\,Q$ if $r$ > $a$. Consequently, the electric field is zero within the sphere and outside module has:

(6:18)

which is an expression identical to the field produced by a charge $Q$concentrated in the center of the sphere. This is the same result found in Appendix B , using a more complicated method (integration).

## Questions

1. A plane with 2500 cm 2 area has a total charge of 20 nC, distributed evenly. The electric field of the module near the plane is approximately:
1. 18.1 mN / C
2. 4:52 kN / C
3. 1.81 N / C
4. 45.2 N / C
5. 0.452 N / C
2. A conductive ball 3 cm radius, isolated and positively charged, produces a field module 36 μN / Nc, which is a point 1 cm from the ball surface. Calculate the total charge of the sphere.
1. 3.6 nC
2. 0.4 nC
3. 1.6 nC
4. 6.4 nC
5. 1.2 nC
3. In a Cartesian coordinate system ($x$, $y$, $z$) (Meters), there is a point charge in two nC (1,0,0), a point charge -4 Nc (0,2,0) and a point charge of 3 nC (0,0, 4). Calculate the electric flux (in SI units) through a sphere of radius 3 m centered at the origin.
1. 36 $\pi$
2. 72 $\pi$
3. -72 $\pi$
4. 108 $\pi$
5. -144 $\pi$
4. The existing charge in a sphere of radius 1 m is distributed in this way unknown. The electric field created by the flux distribution through a radius of the spherical surface 4 m, concentric with the charged sphere, is 11.3 × 10 4  N · m 2 / C. What is the flux of electric field over a distance of 2 m spherical surface concentric with the other two?
1. 45.2 × 10 4  N · m 2 / C
2. 22.6 × 10 4  N · m 2 / C
3. 11.3 × 10 4  N · m 2 / C
4. 56.5 × 10 3  N · m 2 / C
5. 28.2 × 10 3  N · m 2 / C
5. If a closed surface the electric field points into the surface at all points, what can we conclude?
1. There positive charge within the surface.
2. There negative charge in the surface.
3. There is no charge inside the surface.
4. The field is necessarily perpendicular to the surface.
5. The field is necessarily parallel to the surface.

## Problems

1. The atmosphere in an electric field exists which points vertically downwards. The sea level, the module of the field is approximately 120 N / C and decreases as a function of time; 2 km above sea level the field is approximately 66 N / C. What can be concluded about the signal of the free charges in the first two kilometers of the atmosphere? Calculate the average charge density in this region.
2. A point charge is nC 5 to 6 cm in a very long straight wire, a constant linear charge of 7 nC / cm. Calculate the electric force on the wire (hint: better calculate the strength of the wire on the point charge, which is easier to calculate, and the law of action and reaction must have the same module).
3. The figure shows the electric field lines of two charged particles, one at the origin and the other at (1, 0). At the point (3, 0) there is a saddle point. Knowing that the particle charge in the source is 18 nC, calculate the charge of the other particle.
4. Two concentric spherical shells, lightning $a$ and $b$, Lying one inside the other ($a$ < $b$). The inner spherical shell of radius$a$It has a total charge $Q$ uniformly distributed on its surface, and the outer shell has a charge -$Q$Uniformly distributed on its surface. Using Gauss's law to calculate the electric field at points within the inner shell of the two shells and out of the outer shell.
5. A sphere of radius $R$ It has an electric charge $Q$evenly distributed within the volume. Using the law of Gauss, calculate the magnitude of the electric field at a point at a distance$r$the sphere center. Consider the case$r$$R$ and $r$ < $R$.
6. A specific particle mass of 25 g charge of 50 nC is hung from a 7 cm wire that is bonded to a vertical plane. The vertical plane has a constant surface charge $\sigma$= 17 nC / cm 2 and can be considered infinite. Calculate the angle$\theta$ the wire is in the vertical plane.
7. To simulate the charges in a capacitor of parallel flat plates, consider a system 13 charges the value +1 points (-6,5), (-5,5), ..., (5,5), and (6 5) -1 and 13 charges the value of points (-6, -5), (-5, -5), ..., (5, -5) and (6, -5). Using Maxima, draw the electric field and the field lines (the design is independent of the units that are used).

## Anmswers

Questions: 1. B. 2. D. 3. C. 4. C. 5. B.

Problems

1. There is positively charged, with an average charge density equal to 2.39 × 10 -13  C / m 3 .
2. 1:05 mN
3. -8 NC.
4. Inside the inner bark and outside the outer shell, the field is null. Between the two shells, the field is in the radial direction and module$kQ/r^2$, at where $r$ is the distance from the center of the spheres.
5. if $r$$R$, $E=\dfrac{kQ}{r^2}$. if$r$ < $R$, $E=\dfrac{kQ}{R^3}r$
6. 62.99 °
7. Graphic:
Wrong

Calculate the charge per unit area and multiply by 2 π  k for the field module.

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Right

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Wrong

Calculate the charge per unit area and multiply by 2 π  k for the field module.

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Wrong

Units has a problem. Using SI units must pass the area to m 2 and the charge for Coulomb. Using cm 2 and nC, you must convert the value of k , 9 × 10 9 N · m 2 / C 2 , for the units N · cm 2 / nC 2

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Wrong

If you use the value 9 × 10 9 to k , must pass the area cm 2 to m 2 and the charge for Coulomb. The value of the constant k in units of N · cm 2 / nC 2 is not 9.

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Wrong

The distance from the center of the sphere to the point in question is not 4 cm and 3 cm.

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Wrong

The distance from the center of the sphere to the point in question is not 1 cm and 4 cm.

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Wrong

The electric field is equal to k , times the charge, divided by the distance squared .

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Right

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Wrong

The distance from the center of the sphere to the point in question is not 4 cm and 3 cm.

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Wrong

The charges 3 are not all inside the sphere. Only the charges that are produced flux through the interior of the sphere.

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Wrong

The total charge inside the sphere is negative and thus the flux must also be negative.

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Right

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Wrong

Two of the three charges are within the sphere and the total flux through the ball is the sum of these two charge streams.

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Wrong

Two of the three charges are within the sphere and the total flux through the ball is the sum of these two charge streams.

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Wrong

The flux through the two spheres is equal, because the internal charge therein is also the same.

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Wrong

The flux through the two spheres is equal, because the internal charge therein is also the same.

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Right

The flux through the two spheres is equal, because the internal charge therein is also the same.

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Wrong

The flux through the two spheres is equal, because the internal charge therein is also the same.

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Wrong

The flux through the two spheres is equal, because the internal charge therein is also the same.

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Wrong

If the charge were positive, the field would point out.

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Right

The negative flux implies a negative charge.

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Wrong

If there were no charges, the flux through the closed surface would be zero, but as the field always points inside, the flux is nonzero.

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Wrong

Field can point into the surface without being required to be perpendicular to it.

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Wrong

If the field was parallel, would not point neither in nor out of the surface.

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