User:
Particle in electric field of an infinite plate
Concept 
According to Gauss's Law, the variables the surface where the electric field is constant ($dS$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the versor normal to the section ($\hat{n}$), and the electric field ($\vec{E}$) satisfy the following equation:
| $\displaystyle\int_S\vec{E}\cdot\hat{n}\,dS=\displaystyle\frac{Q}{\epsilon_0\epsilon}$ |
In the case of a flat Gaussian surface, the field must be constant, so the relationship of the electric eield ($E$) with the surface of the conductor ($S$) is established as:
| $ E S = \displaystyle\frac{ Q }{ \epsilon_0 \epsilon }$ |
what is shown in the graph
Since the charge density by area ($\sigma$) is equally defined by:
| $ \sigma = \displaystyle\frac{ Q }{ S }$ |
For the electric field of an infinite plate ($E_s$), the resulting expression is:
| $ E_s =\displaystyle\frac{ \sigma }{ 2 \epsilon_0 \epsilon }$ |
ID:(11841, 0)
Particle in electric potential of an infinite plate
Concept
The reference electrical, two infinity plates ($\varphi_d$) is with the electric field, two infinite plates ($E_d$) and the position on the z axis ($z$) is equal to:
| $ \varphi_d = -\displaystyle\int_0^z du\,E_d$ |
the electric field of an infinite plate ($E_s$) is with the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$) and the charge density by area ($\sigma$) is equal to:
| $ E_s =\displaystyle\frac{ \sigma }{ 2 \epsilon_0 \epsilon }$ |
the electric potential, infinite plate ($\varphi_s$) is with and the position on the z axis ($z$) turns out
| $ \varphi_s = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z $ |
As illustrated in the following graph:
the field at two points must have the same energy. Therefore, the variables the charge ($Q$), the particle mass ($m$), the speed 1 ($v_1$), the speed 2 ($v_2$), and the electric potential 1 ($\varphi_1$) according to the equation:
| $ \varphi_1 = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z_1 $ |
and the electric potential 2 ($\varphi_2$), according to the equation:
| $ \varphi_2 = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z_2 $ |
must satisfy the following relationship:
| $ \displaystyle\frac{1}{2} m v_1 ^2 + Q \varphi_1 = \displaystyle\frac{1}{2} m v_2 ^2 + Q \varphi_2 $ |
ID:(11852, 0)
Model
Top
Parameters
Variables
Calculations
to 
, then, select the variable: 
to 
Calculations
Calculations
Variable 
Given Calculate 
Target : Equation To be used 
Equations
$ E_s =\displaystyle\frac{ \sigma }{ 2 \epsilon_0 \epsilon }$
E_s = sigma /(2 * epsilon_0 * epsilon )
$ \displaystyle\frac{1}{2} m v_1 ^2 + Q \varphi_1 = \displaystyle\frac{1}{2} m v_2 ^2 + Q \varphi_2 $
m * v_1 ^2/2 + Q * phi_1 = m * v_2 ^2/2 + Q * phi_2
$ \varphi_s = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z $
phi_s = - sigma * z /( epsilon * epsilon_0 )
$ \varphi_1 = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z_1 $
phi_s = - sigma * z /( epsilon * epsilon_0 )
$ \varphi_2 = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z_2 $
phi_s = - sigma * z /( epsilon * epsilon_0 )
$ \varphi_s = -\displaystyle\int_0^z du\,E_s$
phi_s = -@INT( E_s , u ,0, z )
$ \sigma = \displaystyle\frac{ Q }{ S }$
sigma = Q / S
ID:(15808, 0)
Surface charge density
Equation
The surface charge density is calculated by dividing the total charge by the surface area. Therefore, the relationship between the charge density by area ($\sigma$) and the charge ($Q$) with the surface of the conductor ($S$) is established as:
| $ \sigma = \displaystyle\frac{ Q }{ S }$ |
ID:(11460, 0)
Infinite plate
Equation
The electric field of an infinite plate ($E_s$) is with the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$) and the charge density by area ($\sigma$) equal to:
| $ E_s =\displaystyle\frac{ \sigma }{ 2 \epsilon_0 \epsilon }$ |
According to Gauss's Law, the variables the surface where the electric field is constant ($dS$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the versor normal to the section ($\hat{n}$), and the electric field ($\vec{E}$) satisfy the following equation:
| $\displaystyle\int_S\vec{E}\cdot\hat{n}\,dS=\displaystyle\frac{Q}{\epsilon_0\epsilon}$ |
In the case of a flat Gaussian surface, the field must be constant, so the relationship of the electric eield ($E$) with the surface of the conductor ($S$) is established as:
| $ E S = \displaystyle\frac{ Q }{ \epsilon_0 \epsilon }$ |
Since the charge density by area ($\sigma$) is equally defined by:
| $ \sigma = \displaystyle\frac{ Q }{ S }$ |
For the electric field of an infinite plate ($E_s$), the resulting expression is:
| $ E_s =\displaystyle\frac{ \sigma }{ 2 \epsilon_0 \epsilon }$ |
ID:(11448, 0)
Potential and electric field of a plate
Equation
The electric potential, infinite plate ($\varphi_s$) is with the electric field of an infinite plate ($E_s$) and the position on the z axis ($z$) is equal to:
| $ \varphi_s = -\displaystyle\int_0^z du\,E_s$ |
The base electrical potential ($\varphi_0$) in relation to the electric potential ($\varphi$), the infinitesimal distance ($ds$), and the electric field ($\vec{E}$) is defined by the following equation:
| $ \varphi =\varphi_0 - \displaystyle\int_C \vec{E}\cdot d\vec{s}$ |
Since the field is proportional to distance:
$E_s\propto u$
the simplest path is the distance itself. However, the reference potential cannot be set at infinity because the integral diverges at that point. Therefore, the reference potential must be set at the origin ($u\rightarrow 0$) and can be chosen as zero ($\varphi_0=0$). Thus, the relationship between the electric potential, infinite plate ($\varphi_s$), the electric field of an infinite plate ($E_s$), and the position on the z axis ($z$) is defined by:
| $ \varphi_s = -\displaystyle\int_0^z du\,E_s$ |
ID:(15812, 0)
Electrical potential, surfaces
Equation
The electric potential, infinite plate ($\varphi_s$) is with the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the charge density by area ($\sigma$) and the position on the z axis ($z$) is equal to:
| $ \varphi_s = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z $ |
In the case of an infinite plate, the relationship between the electric potential, infinite plate ($\varphi_s$), the electric field of an infinite plate ($E_s$), and the position on the z axis ($z$) is established by the following equation:
| $ \varphi_s = -\displaystyle\int_0^z du\,E_s$ |
Similarly, the relationship involving the electric field of an infinite plate ($E_s$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the charge density by area ($\sigma$) is defined as:
| $ E_s =\displaystyle\frac{ \sigma }{ 2 \epsilon_0 \epsilon }$ |
In spherical coordinates, this is expressed as:
$\varphi_s = -\displaystyle\int_0^z du \displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon }= -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z$
Finally, the relationship that includes the electric potential, infinite plate ($\varphi_s$) and the position on the z axis ($z$) is determined by the following equation:
| $ \varphi_s = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z $ |
ID:(11586, 0)
Electrical potential, surfaces (1)
Equation
The electric potential, infinite plate ($\varphi_s$) is with the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the charge density by area ($\sigma$) and the position on the z axis ($z$) is equal to:
| $ \varphi_1 = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z_1 $ |
| $ \varphi_s = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z $ |
In the case of an infinite plate, the relationship between the electric potential, infinite plate ($\varphi_s$), the electric field of an infinite plate ($E_s$), and the position on the z axis ($z$) is established by the following equation:
| $ \varphi_s = -\displaystyle\int_0^z du\,E_s$ |
Similarly, the relationship involving the electric field of an infinite plate ($E_s$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the charge density by area ($\sigma$) is defined as:
| $ E_s =\displaystyle\frac{ \sigma }{ 2 \epsilon_0 \epsilon }$ |
In spherical coordinates, this is expressed as:
$\varphi_s = -\displaystyle\int_0^z du \displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon }= -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z$
Finally, the relationship that includes the electric potential, infinite plate ($\varphi_s$) and the position on the z axis ($z$) is determined by the following equation:
| $ \varphi_s = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z $ |
ID:(11586, 1)
Electrical potential, surfaces (2)
Equation
The electric potential, infinite plate ($\varphi_s$) is with the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the charge density by area ($\sigma$) and the position on the z axis ($z$) is equal to:
| $ \varphi_2 = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z_2 $ |
| $ \varphi_s = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z $ |
In the case of an infinite plate, the relationship between the electric potential, infinite plate ($\varphi_s$), the electric field of an infinite plate ($E_s$), and the position on the z axis ($z$) is established by the following equation:
| $ \varphi_s = -\displaystyle\int_0^z du\,E_s$ |
Similarly, the relationship involving the electric field of an infinite plate ($E_s$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the charge density by area ($\sigma$) is defined as:
| $ E_s =\displaystyle\frac{ \sigma }{ 2 \epsilon_0 \epsilon }$ |
In spherical coordinates, this is expressed as:
$\varphi_s = -\displaystyle\int_0^z du \displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon }= -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z$
Finally, the relationship that includes the electric potential, infinite plate ($\varphi_s$) and the position on the z axis ($z$) is determined by the following equation:
| $ \varphi_s = -\displaystyle\frac{ \sigma }{2 \epsilon_0 \epsilon } z $ |
ID:(11586, 2)
Energy of a particle
Equation
Electric potentials, which represent potential energy per unit of charge, influence how the velocity of a particle varies. Consequently, due to the conservation of energy between two points, it follows that in the presence of variables the charge ($Q$), the particle mass ($m$), the speed 1 ($v_1$), the speed 2 ($v_2$), the electric potential 1 ($\varphi_1$), and the electric potential 2 ($\varphi_2$), the following relationship must be satisfied:
| $ \displaystyle\frac{1}{2} m v_1 ^2 + Q \varphi_1 = \displaystyle\frac{1}{2} m v_2 ^2 + Q \varphi_2 $ |
ID:(11596, 0)