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Two plates with opposite charges
Storyboard 
The geometry known as parallel plates can be described as two infinite planes that are electrically charged with equal and opposite charges.
ID:(2076, 0)
Particle in a electric field of a infinite two plate
Concept
In the case of a Gaussian surface for a plane, the electric field ($\vec{E}$) is constant in the direction of the versor normal to the section ($\hat{n}$). Therefore, using the variables the charge ($Q$), the electric field constant ($\epsilon_0$), and the dielectric constant ($\epsilon$), it can be calculated by integrating over the surface where the electric field is constant ($dS$):
| $\displaystyle\int_S\vec{E}\cdot\hat{n}\,dS=\displaystyle\frac{Q}{\epsilon_0\epsilon}$ |
what is shown in the graph
Additionally, the charge density by area ($\sigma$) is calculated using the surface ($S$) and the charge ($Q$) according to the following equation:
| $ \sigma = \displaystyle\frac{ Q }{ S }$ |
Therefore, it follows that the electric field, two infinite plates ($E_d$) is:
| $ E_d =\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon }$ |
ID:(11836, 0)
Particle in a electric potencial of a infinite two plate
Concept
The reference electrical, two infinity plates ($\varphi_d$) in relation to the electric field, two infinite plates ($E_d$) and the position on the z axis ($z$) is given by:
| $ \varphi_d = -\displaystyle\int_0^z du\,E_d$ |
Similarly, the electric field, two infinite plates ($E_d$) in relation to the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the charge density by area ($\sigma$) is defined by:
| $ E_d =\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon }$ |
By integrating from the origin, we obtain:
$\varphi_d = -\displaystyle\int_0^z du \displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon }= -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z$
Thus, the reference electrical, two infinity plates ($\varphi_d$) is given by:
| $ \varphi_d = -\displaystyle\frac{ \sigma }{ \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 }{ \epsilon_0 \epsilon } z_1 $ |
and the electric potential 2 ($\varphi_2$), according to the equation:
| $ \varphi_2 = -\displaystyle\frac{ \sigma }{ \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:(11843, 0)
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Equations
$ E_d =\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon }$
E_d = sigma /( 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_d = -\displaystyle\int_0^z du\,E_d$
phi_d = -@INT( E_d , u ,0, z )
$ \varphi_d = -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z $
phi_d =- sigma * z /( epsilon * epsilon_0 )
$ \varphi_1 = -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z_1 $
phi_d =- sigma * z /( epsilon * epsilon_0 )
$ \varphi_2 = -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z_2 $
phi_d =- sigma * z /( epsilon * epsilon_0 )
$ \sigma = \displaystyle\frac{ Q }{ S }$
sigma = Q / S
ID:(15805, 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)
Two infinite plates with opposite charges
Equation
The electric field, two infinite plates ($E_d$) is with the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$) and the charge density by area ($\sigma$) is equal to:
| $ E_d =\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon }$ |
In the case of a Gaussian surface for a plane, the electric field ($\vec{E}$) is constant in the direction of the versor normal to the section ($\hat{n}$). Therefore, using the variables the charge ($Q$), the electric field constant ($\epsilon_0$), and the dielectric constant ($\epsilon$), it can be calculated by integrating over the surface where the electric field is constant ($dS$):
| $\displaystyle\int_S\vec{E}\cdot\hat{n}\,dS=\displaystyle\frac{Q}{\epsilon_0\epsilon}$ |
Additionally, the charge density by area ($\sigma$) is calculated using the surface ($S$) and the charge ($Q$) according to the following equation:
| $ \sigma = \displaystyle\frac{ Q }{ S }$ |
Therefore, it follows that the electric field, two infinite plates ($E_d$) is:
| $ E_d =\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon }$ |
ID:(11449, 0)
Potential and electric field of two plates
Equation
The reference electrical, two infinity plates ($\varphi_d$) is with the electric field, two infinite plates ($E_d$) and the axle distance ($r$) is equal to:
| $ \varphi_d = -\displaystyle\int_0^z du\,E_d$ |
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:
$E_d\propto u$
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 reference electrical, two infinity plates ($\varphi_d$), the electric field, two infinite plates ($E_d$), and the position on the z axis ($z$) is defined by:
| $ \varphi_d = -\displaystyle\int_0^z du\,E_d$ |
ID:(11578, 0)
Calculation of electrical potentials, double plate
Equation
The reference electrical, two infinity plates ($\varphi_d$) 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_d = -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z $ |
The reference electrical, two infinity plates ($\varphi_d$) in relation to the electric field, two infinite plates ($E_d$) and the position on the z axis ($z$) is given by:
| $ \varphi_d = -\displaystyle\int_0^z du\,E_d$ |
Similarly, the electric field, two infinite plates ($E_d$) in relation to the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the charge density by area ($\sigma$) is defined by:
| $ E_d =\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon }$ |
By integrating from the origin, we obtain:
$\varphi_d = -\displaystyle\int_0^z du \displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon }= -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z$
Thus, the reference electrical, two infinity plates ($\varphi_d$) is given by:
| $ \varphi_d = -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z $ |
ID:(11587, 0)
Calculation of electrical potentials, double plate (1)
Equation
The reference electrical, two infinity plates ($\varphi_d$) 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 }{ \epsilon_0 \epsilon } z_1 $ |
| $ \varphi_d = -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z $ |
The reference electrical, two infinity plates ($\varphi_d$) in relation to the electric field, two infinite plates ($E_d$) and the position on the z axis ($z$) is given by:
| $ \varphi_d = -\displaystyle\int_0^z du\,E_d$ |
Similarly, the electric field, two infinite plates ($E_d$) in relation to the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the charge density by area ($\sigma$) is defined by:
| $ E_d =\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon }$ |
By integrating from the origin, we obtain:
$\varphi_d = -\displaystyle\int_0^z du \displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon }= -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z$
Thus, the reference electrical, two infinity plates ($\varphi_d$) is given by:
| $ \varphi_d = -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z $ |
ID:(11587, 1)
Calculation of electrical potentials, double plate (2)
Equation
The reference electrical, two infinity plates ($\varphi_d$) 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 }{ \epsilon_0 \epsilon } z_2 $ |
| $ \varphi_d = -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z $ |
The reference electrical, two infinity plates ($\varphi_d$) in relation to the electric field, two infinite plates ($E_d$) and the position on the z axis ($z$) is given by:
| $ \varphi_d = -\displaystyle\int_0^z du\,E_d$ |
Similarly, the electric field, two infinite plates ($E_d$) in relation to the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the charge density by area ($\sigma$) is defined by:
| $ E_d =\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon }$ |
By integrating from the origin, we obtain:
$\varphi_d = -\displaystyle\int_0^z du \displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon }= -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z$
Thus, the reference electrical, two infinity plates ($\varphi_d$) is given by:
| $ \varphi_d = -\displaystyle\frac{ \sigma }{ \epsilon_0 \epsilon } z $ |
ID:(11587, 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)