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.

>Model

ID:(2076, 0)


Mechanisms

Description


ID:(15795, 0)


Particle in a electric field of a infinite two plate

Description

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

Description

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

Description



ID:(15805, 0)


Two plates with opposite charges

Description

The geometry known as parallel plates can be described as two infinite planes that are electrically charged with equal and opposite charges.

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$Q$
Q
Charge
C
$\sigma$
sigma
Charge density by area
C/m^2
$\epsilon$
epsilon
Dielectric constant
-
$E_d$
E_d
Electric field, two infinite plates
V/m
$\varphi_1$
phi_1
Electric potential 1
V
$\varphi_2$
phi_2
Electric potential 2
V
$m$
m
Particle mass
kg
$z_1$
z_1
Position at 1
m
$z_2$
z_2
Position at 2
m
$v_1$
v_1
Speed 1
m/s
$v_2$
v_2
Speed 2
m/s
$S$
S
Surface of the conductor
m^2
$q$
q
Test charge
C

Calculations


First, select the equation:   to ,  then, select the variable:   to 

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

 Variable   Given   Calculate   Target :   Equation   To be used



Equations

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)

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)

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)


Examples


(ID 15795)

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)

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)



(ID 15805)

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)

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 }$



(ID 11449)

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 $



(ID 11587)

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 $



(ID 11587)

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)


ID:(2076, 0)