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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)
Model
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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_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)
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)