The Electric Field
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Chapter 1 - The Electric Field

Module 1.1 - Definition

The electric field is actually a particular case of the more extended notion of electromagnetic field, considered only from the point of view of its electric properties.

Associated to the electric field are:

  • the area in which particles experience electric forces and torques and
  • the point function as its characteristic feature.

Module 1.2 - The Electric Field

The electric field in vacuum is studied at a macroscopic level by measuring each point in the field, as well as the magnitude, direction and sense of the force acting upon an electrically charged sample particle.

The test is a particle used to explore the electric field. It has to be as small as possible, so that the force acting upon it may be approximated by the force which would be exerted on it and not by the resultant of forces in the area the particle occupies. For any position in the field, its electric state must be time-invariant.

For the beginning, the study of the electric field is done in vacuum conditions.

Vacuum is a rarefaction limit state meaning a lessening in the particle substance. As a consequence, a point in the vacuum is identified by the vector radii in respect with the origin of a reference system in relative immobility to the particles in the vicinity.

In the electric field, the vacuum force \({\overline F}_{qv}\) exerted upon a particle, depends on the electric charge q and its position in the field spotted by the vectorial radius r.

\[{\overline F}_{qv}={\overline F}_{qv}(q,\;r)\]

By bringing two electrically charged particles with different, but positive charges in the same point P(r) in the electric field, we find that the module ratio equals the charge ratio q 1 and q2.

\[\left|\frac{{\overline F}_{qv}(q_1,\;\overline r)}{{\overline F}_{qv}(q_2,\;\overline r)}\right|=\frac{\left|q_1\right|}{q_2}\]

By bringing the same particle with charge q in two different points P 1(r1) and P2(r2) we find that the module ratio equals the ratio of two functions E v(r1) and Ev(r2).

\[\frac{\left|{\overline F}_{qv}(r_1,\;q)\right|}{\left|{\overline F}_{qv}(r_2,\;q)\right|}=\frac{\left|E_v(r_1)\right|}{\left|E_v(r_2)\right|}\]

Based on the previous relations, which of the following formulas will be derived:

\({\overline F}_{qv}(q,r)=r{\overline E}_v(q)\)
\({\overline F}_{qv}(q,r)={\overline E}_v(r)\)
\({\overline F}_{qv}(q,r)={\overline E}_v(r,q)\)
Check
DEFINITION

The vectorial magnitude Ev(r) equals the ratio between the force \({\overline F}_{qv}(r)\) exerted in vacuum over the particle by its electric charge q, when this vanishes, is a local characteristic of the electric field in vacuum and is called the strength of the electric field in vacuum.

\[{\overline E}_v(r)=\lim_{q\rightarrow0}\frac{\overline F(q,\;\overline r)}q\]

In SI, Ev is marked [V/m] and is the electric field vector in which a force of 1N is exerted upon the particle of charge 1 C.

DEFINITION

The field line is a curve for which in each point the vector \({\overline E}_{v}\) is oriented as the tangent is, and in the same direction.

Since in each point the vector \({\overline E}_{v}\) is uniquely determined, the field lines do not cross.

Denoting by ds the length of the field line oriented in its direction, the differential equation of the field line is the following:

\[d\overline s\times{\overline E}_v=0\]
DEFINITION

A homogenous or uniform electric field is the one in which each vectorial point E v has the same value and orientation, the field lines being parallel and equi-spaced.

Chapter 2 - The Electric Charge

Module 2.1 - Definition

DEFINITION

The Electric Charge is defined as the force exerted in an even electric field of strength E v upon the inserted particle.

Based on the definition presented above, indicate which one of the assertions bellow is true and which one is false:

The correct formula for the electric charge is \(q=\frac{{\overline E}_v}{{\overline F}_q}\)

True
False

The correct formula for the electric charge is \(q=\frac{{\overline F}_q}{{\overline E}_v}\)

True
False
Check

If in a time-varying physical system the electric charges q1, q2, q3... satisfy the condition that at each moment their sum is nil,

\[\sum_{k=1}^nq_n=0\]

then they make up a complete system of electric charges. If their sum is non-nil they make up an incomplete system.

Module 2.2 - Charge Distribution

We define:

  • Volume density \[\rho_v=\lim_{\triangle_v\rightarrow0}\frac{\triangle q}{\triangle V}=\frac{dq}{dV'}\]
  • Surface density \[\rho_A=\lim_{\triangle A'\rightarrow0}\frac{\triangle q}{\triangle A'}=\frac{dq}{dA'}\]
  • Linear density

Chapter 3 – Coulomb's Law

Module 3.1 - Definition

Charles Coulomb (1736–1806) measured the magnitudes of the electric forces between charged objects using the torsion balance, which he invented (see the video below).

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Coulomb's electrical torsion balance

In the image above, given the q1 and q2 electric charges in two particles in vacuum situates at the distance R, forces F12 and F21 exerting over particles 1 and 2 have the following properties:

Choose the correct property in each of the following 3 statements from the drop-down lists:
  • satisfy the principle of action and reaction:
    choose...
    • F12 = - F21
    • F12 = 2F21
    • 2F12 = F21
  • Choose...
    • like
    • opposite
    charges repel and
    choose...
    • like
    • opposite
    charges attract
  • depending on the electric properties of the vacuum, under absolute value, forces are
    choose...
    • proportional
    • inversely proportional
    to the charge product and
    choose...
    • proportional
    • inversely proportional
    to the square distance
Check

The resulting formula is: \[F_{12}=F_{21}=\frac1{4{\mathrm{πε}}_0}\cdot\frac{q_1q_2}{R^2}\] where \({\mathrm\varepsilon}_0\) is the vacuum permittivity [F/m].

Considering R = lm , q1 = q2 = lC and F 12 = F21 = 9 - 1O9 N, the correct formula for \(\varepsilon_0\) is:
\(\varepsilon_0=36\pi\cdot10^9-\left[\frac{C^2}{Nm^2}\right]\)
\(\varepsilon_0=\frac{36\pi}{10^9}-\left[\frac{C^2}{Nm^2}\right]\)
\(\varepsilon_0=\frac1{36\pi\cdot10^9}-\left[\frac{C^2}{Nm^2}\right]\)
Check

Module 3.2 - The Coulombian Electric Field

According to the formula \({\overline F}_{qv_T}=q{\overline E}_v\), force F12 equals the product of electric charge q1 and the vector electric field in vacuum \({\overline E}_{12}\), established by charge q2.

\({\overrightarrow F}_{12}=q_1{\overrightarrow E}_{12}\)
\({\overrightarrow F}_{21}=q_2{\overrightarrow E}_{21}\)

Thus:

\({\overline E}_{12}=\frac1{4{\mathrm{πε}}_0}\cdot\frac{q_1}{R^2}{\overline u}_{21}\)
\({\overline E}_{21}=\frac1{4{\mathrm{πε}}_0}\cdot\frac{q_2}{R^2}{\overline u}_{12}\)

A point charge q, establishes in a certain point P of distance R an electrostatic field whose field vector Ev is radial, in proportion to the charge q and inversely proportional to the square distance.

\[{\overline E}_v=\frac1{4{\mathrm{πε}}_0}\cdot\frac q{R^2}{\overrightarrow u}_R=\frac q{4{\mathrm{πε}}_0}\cdot\frac{\overrightarrow R}{R^3}\]

The strength of the electrostatic field is oriented outward from the particle in case of a positive charge and inward in case of a negative charge.

In order to deduce the formula for \({\overrightarrow E}_v\), assess the reasoning within the three demonstrations bellow and indicate which one is correct.
The strength of the electrostatic field Ev established in a point in vacuum of n charges qn, is the vectorial sum of the strengths due to all point charges.
\[{\overline E}_v=\underset{}{\sum{\overline E}_{vk}}\]
where \({\overline E}_{vk}=\frac{q_k}{4\pi\varepsilon_0}\cdot\frac{{\overline R}_k}{R_k^3}\)
thus
\[{\overrightarrow E}_v=4\pi\varepsilon_0\underset{}{\sum\frac{q_k}{R_k^3}\cdot{\overline R}_k}\]
Correct
Incorrect
The strength of the electrostatic field Ev established in a point in vacuum of n charges qn, is the vectorial sum of the strengths due to all point charges.
\[{\overline E}_v=\underset{}{\sum{\overline E}_{vk}}\]
where \({\overline E}_{vk}=\frac{q_k}{4\pi\varepsilon_0}\cdot\frac{{\overline R}_k}{R_k^3}\)
thus
\[{\overrightarrow E}_v=\frac1{4\pi\varepsilon_0}\underset{}{\sum\frac{q_k}{R_k^3{\overline R}_k}}\]
Correct
Incorrect
The strength of the electrostatic field Ev established in a point in vacuum of n charges qn, is the vectorial sum of the strengths due to all point charges.
\[{\overline E}_v=\underset{}{\sum{\overline E}_{vk}}\]
where \({\overline E}_{vk}=\frac{q_k}{4\pi\varepsilon_0}\cdot\frac{{\overline R}_k}{R_k^3}\)
thus
\[{\overrightarrow E}_v=\frac1{4\pi\varepsilon_0}\underset{}{\sum\frac{q_k}{R_k^3}\cdot{\overline R}_k}\]
Correct
Incorrect
Check

If charge q is distributed, the strength of the elementary electrostatic field Ev established in vacuum by the elementary charge dq is computed by:

\[d{\overline E}_v=\frac{dq}{4{\mathrm{πε}}_0}\cdot\frac{\overline R}{R^3}\]

If the charge has a volumetric distribution of density \(\rho_v(r')\), of superficial surface \(\rho_A(r')\) or linear \(\rho_1(r')\) we can consider that:

\[dqv=\rho_v(r')dv'\]
\[dqA=\rho_A(r')dA'\]
\[dq_1=\rho_1(r')ds'\]

and vector \({\overline E}_{v}\) we integrate on volume V, surface S and line C.

If there are discrete charges in the field, the strength of the electrostatic field is computed by the following:

\[{\overline E}_v=\frac1{4{\mathrm{πε}}_0}\cdot\frac{\overline R}{R^3}\]
\[ \begin{align} {\overline E}_v(r) = \frac1{4\pi\varepsilon_0}\cdot\left[\int_v\rho_v(r')\cdot\frac{\overline R}{R^3}dv +\int_A\rho_A(r')\cdot\frac{\overline R}{R^3}dA' +\int_C\rho_1(r')\cdot\frac{\overline R}{R^3}ds' +\sum_{}\frac{q_k}{R_k^3}{\overline R}_k\right] \end{align} \]
\[ \begin{align} {\overline E}_v(r) = \frac1{4\pi\varepsilon_0}\cdot\left[\int_v\rho_v(r')\cdot\frac{\overline R}{R^3}dv +\int_A\rho_A(r')\cdot\frac{\overline R}{R^3}dA'\\ +\int_C\rho_1(r')\cdot\frac{\overline R}{R^3}ds' +\sum_{}\frac{q_k}{R_k^3}{\overline R}_k\right] \end{align} \]
\[ \begin{align} {\overline E}_v(r) = \frac1{4\pi\varepsilon_0}\cdot\left[\int_v\rho_v(r')\cdot\frac{\overline R}{R^3}dv\\ +\int_A\rho_A(r')\cdot\frac{\overline R}{R^3}dA'\\ +\int_C\rho_1(r')\cdot\frac{\overline R}{R^3}ds'\\ +\sum_{}\frac{q_k}{R_k^3}{\overline R}_k\right] \end{align} \]

Elements of volume, surface, length are given by vector radius \(\overline r'\) and point P by radius \(\overline r\).

The distance between point P and P' is:

\[ \begin{split} \left|\overline R\right| =& \left|\overline r-\overline r'\right|\\ =& \left|(x-x')\overline i+(y-y')\overline j+(z-z')\overline k\right| \end{split} \]
\[ \begin{split} \left|\overline R'\right| =& \left|\overline r'-\overline r\right|\\ =& \left|(x'-x)\overline i+(y'-y)\overline j+(z'-z)\overline k\right| \end{split} \]

and

\[\frac{\partial R}{\partial x}=-\frac{\partial R'}{\partial x'}\]
\[\nabla=\overline i\frac\partial{\partial x}+\overline j\frac\partial{\partial y}+\overline k\frac\partial{\partial z}\]
\[\nabla'=\overline i\frac\partial{\partial x'}+\overline j\frac\partial{\partial y'}+\overline k\frac\partial{\partial z'}\]

and \(\nabla=\nabla'\)

If U(R) is a scalar function and \(\overline F(R)\) a vector, then:

  • gradU(R) = - grad'U(R)
  • dirF(R) = - div'F(R)
  • rotF(R) = - rot'F(R)

Since \(\frac{\overline R}{R^3}=-grad\frac1R\)

The defining relation of the electrostatic field strength then becomes: \[{\overline E}_v(r')=-\frac q{4{\mathrm{πε}}_0}grad\frac1R\]

Interpreting the formula presented above, which would be the correct statement for \({\overline E}_v(r')\):
The strength of the electrostatic field established by a particle charge in a point at a distance R is proportional to the gradient of distance R.
The strength of the electrostatic field established by a particle charge in a point at a distance R is proportional to the gradient of inverse distance R.
The strength of the electrostatic field established by a particle charge in a point at a distance R is proportional to the inverse distance R.
Check

Chapter 4 – The Electric Potential

Module 4.1 - Definition

DEFINITION

We call electric potential u12 the scalar magnitude equal to the linear integral of the scalar product between the strength of the electric field \({\overline E}_v\) and length ds between points P1 and P2 of curve \(\Gamma\).

Under static and stationary regimes, the electric potential does not vary in time U12. If the electric potential is time-varying it is denoted with u12 and is called instantaneous.

The electric potential globally characterises the electric field referring to a given curve \(\Gamma\) between two of its points.

The integration sense, i.e. the sense of the curve arc element \(\Gamma\), is called the reference sense from P1 to P2 if point P spotted by the vector radius r, length \(d\overline s\) is assimilated to the differential of vector radius \(d\overline r\).

\[d\overline s=d\overline r\]
\[u_{12}=\int_{1\Gamma}^2{\overline E}_vd\overline r\]

In an even electric field the electric potential u12 between two points P1 and P 2 at distance d is not dependent on the curve \(\Gamma\) shape and has the following relation:

\[u_{12}=\int_{1\Gamma}^2{\overline E}_vd\overline s=E_v\int_{1\Gamma}^2ds\;\cos\alpha=E_vd\;\cos\alpha\]

If points P1 and P2 are on the same field line and \(\alpha=0\), the electric potential is positive and \(u=E_vd>0\). What will this lead to?

Taking into account the premises presented above, indicate if each of the following three conclusions are true or false:

The vector's module Ev is equal to the drop in field line on the length unit of the electric potential.

True
False

The vector's module Ev is equal to the length unit of the field line on the drop in electric potential.

True
False

The vector's module Ev is equal to the drop in electric potential on the length unit of the field line.

True
False
Check

If points P1 and P2 are on a line perpendicular on the field lines \(\alpha=\pm\mathrm\pi/2\), then the electric potential is identical nil.

If \({\overrightarrow{\mathrm E}}_\mathrm v\) has in rectangular coordinates the components Ex, Ey, Ez and ds element components dx, dy, dz, electric potential is computed by the relation below:

\[\overrightarrow E=Ex\overrightarrow i+Ey\overrightarrow j+Ez\overrightarrow k\]
\[ds=dx\overrightarrow i+dy\overrightarrow j+dz\overrightarrow k\]
\[u_{12}=\int_1^2(E_xdx+E_ydy+E_zdz)\]
DEFINITION

The line integral of the electric field strength exerted on a closed curve \(\Gamma\) is called electromotive potential and is denoted with er.

\[e_r=\oint_\Gamma{\overrightarrow E}_vd\overline s\]

In a uniform field, the electromotive potential is nil for any closed curve \(\Gamma\).

Module 4.2 - Electrostatic potential; Potential difference

Let us consider a point charge q in the reference origin. The electric potential U12 between points P 1 and P2 of curve \(\Gamma\) is computed by:

\[u=\int_{1\Gamma}^2{\overline E}_vd\overline s\]

Ev(r) is the strength of the electrostatic field in a point P on the curve computed by the relation:

\[{\overline E}_v(r)=\frac q{4\pi\varepsilon_0}\cdot\frac{\overline R}{R^3}\]

Replacing

\[ \begin{align} U_{12} &= \int_{1\Gamma}^2\frac q{4\pi\varepsilon_0}\cdot\frac{\overline rd\overline s}{r^3}\\ &= \int_{1\Gamma}^2\frac q{4\pi\varepsilon_0}\cdot\frac{ds\;\cos\alpha}{r^2}\\ &= \frac q{4\pi\varepsilon_0}\cdot\left(\frac1r\right)\mbox{\large$/$}_1^2\\ &= \frac q{4\pi\varepsilon_0}\cdot\left(\frac1{r_1}-\frac1{r_2}\right) \end{align} \]

function V is defined by the relation: \(V=\frac q{4\pi\varepsilon_0}\cdot\frac1r\) and U12=V1-V2 where \(V_1=\frac q{4\pi\varepsilon_0}\cdot\frac1{r_1}\); \(V_2=\frac q{4\pi\varepsilon_0}\cdot\frac1{r_2}\)

If point P2 tends to infinite, the electric potential \(U_{1\infty_{r_2\rightarrow0}}=\lim_{}U_{12}=V_1=\frac q{4\pi\varepsilon_0r_1}\)

The line integral of the electrostatic field on a curve of a certain shape, between point P of distance r from the charge and the point in the infinite is called electrostatic potential V.

On considering P1 a current point P and P2 a reference point P0, potential P is computed by the following relation:

\[V_P=V_0-\int_0^P{\overline E}_v\cdot d\overline s\]
\[U_{12}=V_1-V_2\]
\[V_2=V_1-U_{12}=V_0-\int_0^P{\overline E}_v\cdot d\overline s\]

The electric potential U12 equal to the difference between potentials V1 and V 2 is called potential difference.

The electrostatic potential V(r) is a continuous point function except for singularities. The choice of reference point P0 whose reference potential V0 enters the potential relation V P is random, provided that the integral of the electric field between points P0 and P1 does not have infinite values. In this case, the point P0 is not specified and V0 is an additive constant.

\[V=\int_{}^P{\overline E}_v\cdot d\overline s+const\]

The electrostatic potential V is defined according to the definition of the potential energy of a material point:

  • Vector Ev is the Coulombian force \({\overline F}_{qv}\) exerted over the point charge unit
  • The potential difference is the work L12 from outside needed to move the test particle with unit electric charge from point P 1 to P2.
\[L_{12}=\int_{P_1}^{P_2}{\overrightarrow F}_{qv}\cdot d\overline s\]
\[L_{12}=\int_{P_1}^{P_2}{\overrightarrow F}_{qv}\cdot d\overline s=V_1-V_2\]

Module 4.3 – Potential Gradient

If we differentiate the relation \(V=V_0-\int_0^P{\overline E}_v\cdot d\overline s\) we get \(dV=gradV\cdot d\overline s\)

By comparing the two relations, we get: \(E_v=-gradV=-\Delta V\)

It results that the strength of the electric field Ev is the gradient with the changed sign of potential V.

Module 4.4 - The Electrostatic Potential of Distributed Charges

According to the superposition principle, the electrostatic potential V established in a certain point, in the vacuum, of n charges qk is equal to the sum of the electrostatic potentials V k that each charge would have established in that point.

In order to deduce the formula for V, assess the reasoning within the three demonstrations bellow and indicate which one is correct.
Since \(V=\sum_{k=1}^nV_k\)
and \(V_k=\frac{q_k}{4\pi}\cdot\frac{\varepsilon_0}{R_k}\)
then
\[V=\frac{\varepsilon_0}{4\pi}\cdot\sum_{k=1}^n\frac{q_k}{R_k}\]
Correct
Incorrect
Since \(V=\sum_{k=1}^nV_k\)
and \(V_k=\frac{q_k}{4\pi\varepsilon_0}\cdot\frac1{R_k}\)
then
\[V=\frac1{4\pi\varepsilon_0}\cdot\sum_{k=1}^n\frac{q_k}{R_k}\]
Correct
Incorrect
Since \(V=\sum_{k=1}^nV_k\)
and \(V_k=\frac{R_k}{4\pi\varepsilon_0}\cdot\frac1{q_k}\)
then
\[V=\frac1{4\pi\varepsilon_0}\cdot\sum_{k=1}^n\frac{q_k}{R_k}\]
Correct
Incorrect
Check

For a distributed charge, \(dV=\frac{dq}{4\pi\varepsilon_0}\cdot\frac1r\).

Module 4.5 - Equipotential Surface

Since the electrostatic potential is a scalar potential function, we can draw surfaces whose points have the same potential, called equipotential surfaces.

V(r) = V(x,y,z) = const.

Since the strength of the electric field is the gradient of opposed sign of the potential, the field lines of the electric field are perpendicular on equipotential surfaces.

Chapter 5 – The Electric Potential

Module 5.1 – Definition of the Electric Field Flux

DEFINITION

The vectorial magnitude equal to the product between the strength of the electric field in vacuum \({\overrightarrow E}_v\) and permittivity \(\varepsilon_0\) is called electric inductivity in vacuum \({\overrightarrow D}_v\).

\[{\overrightarrow D}_v=\varepsilon_0\cdot{\overrightarrow E}_v\]
DEFINITION

The electric flux is the scalar magnitude equal to the surface integral of the scalar product between the electric induction and surface.

\(\Psi_s=\int_{S_\Gamma}{\overrightarrow D}_v\cdot\overrightarrow n\;dA\) for an outward facing surface \(\Psi_s\)

\(\Psi_\Sigma=\int_\Sigma{\overrightarrow D}_v\cdot\overrightarrow n\;dA\) for an inward facing surface

Module 5.2 – Gauss’s Law

We consider an outward facing surface \({\overrightarrow S}_\Gamma\) on \(\Gamma\), which has a coverage direction.

The electric inductivity in vacuum \({\overrightarrow D}_v\), in a point P on surface \({\overrightarrow S}_\Gamma\) is computed by the following relation:

\[{\overrightarrow D}_v=\varepsilon_0{\overrightarrow E}_v=\frac q{4\mathrm\pi}\cdot\frac{\overline R}{R^3}\]

And for the electric flux \(\Psi_{S_\Gamma}\) we get:

\[\Psi_{S_\Gamma}=\int_{S_\Gamma}{\overrightarrow D}_v\cdot\overrightarrow n\;dA=\frac q{4\mathrm\pi}\int_{S_\Gamma}\frac{\overrightarrow R\cdot\overrightarrow n}{R^3}dA\]
\[\Psi_{S_\Gamma}=\frac q{4\mathrm\pi}\int_{S_\Gamma}\frac{dA\;\cos\alpha}{R^2}\]

magnitude \(\int_{S_\Gamma}\frac{dA\;\cos\alpha}{R^2}=\Omega_\Gamma\) is the solid angle under which we may see the curve in the point where the charge is located.

If there are n charges in the field, according to the superposition principle, the electric flux \(\Psi_{S_\Gamma}\) has the following expression:

\[\Psi_{S_\Gamma}=\frac1{4\mathrm\pi}\sum_{k=1}^nq_k\Omega_{\Gamma,k}\]

For closed surfaces, if we consider the charge enclosed, the solid angle in a point inside the closed surface being equal to \(4\mathrm\pi\), we get \(\Psi_\Sigma=q\)

This result may be explained as follows:

  • the cone with the vertex in which is the charge whose generators are tangent to the surface \(\Sigma\), determine a curve \(\Gamma\) which separates two open surfaces \(S\Gamma\) and \(S'\Gamma\) \((S\Gamma\cup S'\Gamma=\Sigma)\).

Considering at random a coverage direction for curve \(\Gamma\), the normal versor of one of the open surfaces is identical to the versor of surface \(\Sigma\) and the versor of the other has an opposed direction.

\[n_{S_\Gamma}=n_\Sigma\]
\[n_{S'_\Gamma}=-n_\Sigma\]

Therefore \(d\Omega_{S_\Gamma}=-d\Omega'_{S'_\Gamma}\) and \(\Omega_\Sigma=\Omega_{S_\Gamma}+\Omega'_{S'_\Gamma}=0\).

DEFINITION

Considering the above, the theorem of the electric flux (or Gauss's Law) can be expressed as follows:

The electric flux in any closed surface \(\Sigma\) drawn exclusively in vacuum is equal to the electric charge \(q_\Sigma\) inside it.

Chapter 6 – Application

The field and the electric potential in case of a plane S evenly charged with surface density \(\rho\) of electric charges.

For symmetrical reasons, \({\overrightarrow E}_v\) is \(\perp\) on the plane.

Thus, the flux on the lateral surface is "0".

\[{\overrightarrow E}_v\cdot dA=0\]

The flux is reduced to its value by surfaces S and S' of equal areas, A.

\[\Psi_E=\int_{S_1}{\overline E}_v\cdot d\overline A+\int_{S_2}{\overline E}_v\cdot d\overline A=2AE_V\;\;\;\;\;\;(A\perp E_V)\]
\[\Psi_E=\int_{S_1}{\overline E}_v\cdot d\overline A+\int_{S_2}{\overline E}_v\cdot d\overline A=2AE_V\]
\[(A\perp E_V)\]

From GAUSS's Law:

\[\Psi_E=\frac1{\varepsilon_0}q=\frac1{\varepsilon_0}A\rho_S\]

Equalling the two formulas above, indicate which of the following formulas will be obtained:

\({\overrightarrow E}_v=\frac{\rho_S}{2\varepsilon_0}\) and \({\overrightarrow E}_v=\frac{\rho_S}{2\varepsilon_0}\overline k=\overrightarrow{E_v^{'}}\)
\({\overrightarrow E}_v=\frac{\rho_S}{2\varepsilon_0}\) and \({\overrightarrow E}_v=\frac{\rho_S}{2\varepsilon_0}\overline k=-\overrightarrow{E_v^{'}}\)
\({\overrightarrow E}_v=\frac{\rho_S}{2\varepsilon_0}\) and \({\overrightarrow E}_v=-\frac{\rho_S}{2\varepsilon_0}\overline k=-\overrightarrow{E_v^{'}}\)
Check