(ID 15798)

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)

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)

(ID 15808)

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 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 }$

(ID 11448)

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 $

(ID 11586)

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 $

(ID 11586)

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)